MECHANICS 


A  TEXT-BOOK   FOR   ENGINEERS 


BY 

JAMES  E.  BOYD,  M.S. 

PROFESSOK  OF  MECHANICS,  THE  OHIO  STATE  UNIVERSITY 
AUTHOR  OF  "STRENGTH  OF  MATERIALS,"  "DIFFERENTIAL  EQUATIONS  FOR 

ENGINEERING   STUDENTS,"   ETC. 


FIRST  EDITION 


McGRAW-HILL  BOOK  COMPANY,  INC. 

NEW  YORK:    370  SEVENTH  AVENUE 

LONDON:    6*8  BOUVERIE  ST.,  E.  C.  4 

1921 


COPYRIGHT,  1921,  BY  THE 
McGRAW-HiLL  BOOK  COMPANY,  INC. 


THE    MAFL  E    FXK  8  S    T  O  H  K    PA. 


PREFACE 

This  book  is  intended  to  give  a  working  knowledge  of  the 
principles  of  Mechanics  and  to  supply  a  foundation  upon  which 
intelligent  study  of  Strength  of  Materials,  Stresses  in  Structures, 
Machine  Design,  and  other  courses  of  more  technical  nature  may 
rest. 

In  the  development  of  this  subject,  emphasis  is  put  upon  the 
physical  character  of  the  ideas  involved,  while  mathematics  is 
employed  as  a  convenient  tool  for  the  determination  and  expres- 
sion of  quantitative  relations.  Analytical  and  graphical  methods 
are  given  together  and  each  is  interpreted  in  terms  of  the  other. 
While  the  principal  stress  is  placed  upon  Mechanics  as  a  science, 
considerable  attention  is  given  to  Mechanics  as  an  art.  In  the 
text,  in  some  of  the  problems,  and  in  many  of  the  illustrative 
examples,  methods  of  calculation  are  suggested  by  means  of 
which  accurate  results  may  be  most  readily  obtained. 

The  definitions  of  work  and  potential  energy,  together  with  the 
solution  of  problems  of  statics  by  the  method  of  virtual  work 
are  given  early.  In  the  treatment  of  dynamics,  the  definitions 
of  kinetic  energy  and  its  application  to  the  conditions  of  variable 
motion  are  introduced  as  soon  as  possible. 

In  equations  involving  acceleration  or  energy,  the  common 
commercial  units  are  employed — the  pound  mass  as  the  unit  of 
mass  and  the  weight  of  the  pound  mass  as  the  unit  of  force.  In 
order  to  clear  up  the  confusion  which  results  from  the  fact  that 
physicists  use  one  set  of  units  while  some  engineering  writers 
use  another,  Chapter  XVIII  is  devoted  to  a  discussion  of  the 
various  systems. 

The  author  acknowledges  his  obligations  to  many  of  his  col- 
leagues who  have  assisted  in  the  preparation  of  this  book.  P. 
W.  Ott  of  the  Department  of  Mechanics  checked  the  problems 
of  several  chapters.  S.  A.  Harbarger  of  the  Department  of 
English  read  all  the  manuscript  and  assisted  in  the  final  revision. 
Professor  O.  E.  Williams  of  the  Department  of  Engineering 


vi  PREFACE 

Drawing  supervised  the  preparation  of  the  drawings,  and 
Professor  Robert  Meiklejohn  of  the  same  department  made 
many  valuable  suggestions. 

J.  E.  BOYD. 
COLUMBUS,  OHIO, 
May  13,  1921. 


CONTENTS 

PAGE 

PREFACE v 

NOTATION xiii 

CHAPTER  I 

FUNDAMENTAL  IDEAS 1 

Mechanics — Illustrations — Fundamental      Quantities — Standards 
and  Units — Length — Time — Matter  and  Mass — Force — Weight — 
yX  Relation  of  Mass  to  Weight — Variation  of  Weight  with  Latitude 
and  Altitude. 

CHAPTER  II 

QUANTITY  AND  CALCULATIONS 9 

Representation  of  Quantity  by  Numbers — Representation  of 
Quantity  by  Lines-^Vectors— ^Addition  of  Vectors-vComponents 
of  a  Vector — Computation  of  the  Vector  Sum — Vector  Difference — 
Vectors  in  Space — Vectors  by  Direction  Cosines — Addition  of 
Vectors  in  Space — Graphical  Resolution  of  Vectors  in  Space — 
Graphical  Determination  of  the  Vector  Sum — Product  of  Two 
Vectors — Summary. 

CHAPTER  III 

APPLICATION  OF  FORCE  .    .   .   .• 29 

y  Tension,  Compression,  and  Shear — States  of  Matter — 4  Rigid 
Body — A  Flexible  Cord — Eguilibrium — A  smooth  Surface — A 
Smooth  Hinge — Resultant  and  Equilibrant — The  Force  Circuit — 
The  Free  Body — Application  of  Forces — Resultant  of  Concurrent 
Forces — Resolution  of  Forces — Work — Classes  of  Equilibrium. 

CHAPTER  IV 

CONCURRENT  CO-PLANAR  FORCES 43 

/>(  Resultant — Calculation  of  Resultants  and  Components — Equi- 
librium— Equilibrium  by  Resolutions — Trigonometric  Solution — 
Number  of  Unknowns — Moment  of  a  Force — Moment  of  Resultant 
— Equilibrium  by  Moments — Conditions  for  Independent  Equa- 
tions— Connected  Bodies — Bow's  Method — Summary — Miscellane- 
ous Problems. 

CHAPTER  V 

NON-CONCURRENT  CO-PLANAR  FORCES 72 

Resultant  of  Parallel  Forces  in  the  Same  Direction — Resultant  of 
Parallel  Forces  in  Opposite  Directions — Equilibrium  of  Parallel 
Co-planar  Forces — Condition  of  Stable  Equilibrium — Resultant  of 

vii 


viii  CONTENTS 

PAGE 

Non-parallel  Forces  Graphically — Calculation  of  the  Resultant  of 
Non-parallel  Forces — Equilibrium  of  Non-concurrent  Forces — 
Condition  for  Independent  Equations — Direction  Condition  of 
Equilibrium — Trusses — Method  of  Sections — Jointed  Frame  with 
Non-concurrent  Forces — Summary — Miscellaneous  Problems. 

CHAPTER  VI 

..  COUPLES Ill 

Moment  of  a  Couple — Equivalent  Couples — Algebraic  Addition 
of  Couples — Equilibrium  of  Couples — Reduction  of  a  Force  and  a 
Couple  to  a  Single  Force — Resolution  of  a  Force  into  a  Force  and  a 
Couple — Summary. 

CHAPTER  VII 

GRAPHICS  OF  NON-CURRENT  FORCES 120 

Resultant  of  Parallel  Forces — Resultant  of  Non-parallel  Forces 
— Parallel  Reactions — Non-parallel  Reactions. 

CHAPTER  VIII 

FLEXIBLE  CORDS 129 

The  Catenary — Deflection  in  Terms  of  the  Length — Deflection  in 
Terms  of  the  Span — Solution  by  Infinite  Series — Cable  Uni- 
formly Loaded  per  Unit  of  Horizontal  Distance — Summary — 
Miscellaneous  Problems. 

CHAPTER  IX 

y     CONCURRENT  NON-COPLANAR  FORCES 141 

Resultant  and  Components — Resolution  of  Non-coplanar  Forces — 
Calculation  of  Resultant — Equilibrium  of  Concurrent,  Non- 
coplanar  Forces — Moment  About  an  Axis — Equilibrium  by 
Moments — Summary. 

CHAPTER  X 

..  PARALLEL  FORCES  AND  CENTER  OF  GRAVITY 149 

Resultant  of  Parallel  Forces — Equilibrium  of  Parallel  Forces — 
Center  of  Gravity — Center  of  Gravity  Geometrically — Center  of 
Gravity  of  a  Triangular  Plate — Center  of  Gravity  of  a  Pyramid — 
Center  of  Gravity  of  Integration — Combination  Methods  of  Calcu- 
lation— Center  of  Gravity  of  Some  Plane  Areas — Liquid  Pressure 
— Center  of  Pressure — Summary. 

CHAPTER  XI 

y     FORCES  IN  ANY  POSITION  AND  DIRECTION 173 

Couples  in  Parallel  Planes — A  Couple  as  a  Vector — Resultant 
Couple — Forces  reduced  to  a  Force  and  Couple — Equilibrium  of 
Non-concurrent,  Non-coplanar  Forces — Summary. 


CONTENTS  ix 

PAGE 

CHAPTER  XII 

FRICTION 189 

Coefficient  of  Friction — Angle  of  Friction — Cone  of  Friction — 
Bearing  Friction — Rolling  Friction — Roller  Bearings — Belt 
Friction — Summary — Miscellaneous  Problems. 

CHAPTER  XIII 

WORK  AND  .MACHINES „ 205 

Work  and  Energy — Equilibrium  by  Work — Machines — Inclined 
Plane— Wheel  and  Axle— Pulleys— The  Screw— Differential 
Applicances — The  Lever — Virtual  Work — Character  of  Equi- 
librium— The  Beam  Balance — Surface  of  Equilibrium — Summary. 

CHAPTER  XIV 

MOMENT  OF  INERTIA  OF  SOLIDS 228 

Definition — Moment  of  Inertia  by  Integration — Radius  of 
Gyration — Transfer  of  Axis — Moment  of  Inertia  of  a  Thin  Plate 
— Plate  Elements — Moment  of  Inertia  of  Connected  Bodies — 
Summary — Miscellaneous  Problems. 

CHAPTER  XV 

y  MOMENT  OF  INERTIA  OF  A  PLANE  AREA 241 

Definition — Polar  Moment  of  Inertia — Axis  in  Plane — Relation 
of  Moments  of  Inertia — Product  of  Inertia — Transfer  of  Axes  for 
Product  of  Inertia — Change  of  Direction  of  Axis  for  Moment  of 
Inertia — Change  of  Direction  of  Axes  for  Product  of  Inertia — 
Maximum  Moment  of  Inertia — Center  of  Pressure — Moment  of 
Inertia  by  Moment  of  a  Mass — Summary. 

CHAPTER  XVI 

MOTION 259 

Displacement — Velocity — Average  and  Actual  Velocity — Velocity 
as  a  Derivative — Acceleration — Acceleration  as  a  Derivative — 
Acceleration  as  a  Vector — Dimensional  Equations — Summary. 

CHAPTER  XVII 

FORCE  AND  MOTION 273 

Force  and  Acceleration — Constant  Force — Integration  Methods 
— Connected  Bodies — Velocity  and  Displacement — Energy — 
— Potential  Energy — Motion  Due  to  Gravity — Projectiles — 
Summary — Miscellaneous  Problems. 

CHAPTER  XVIII 

SYSTEM  OF  UNITS 294 

Gravitational  System — Absolute  Systems — The  Centimeter — 
Gram — Second  System — The  Foot — Pound — Second  System — 
The  Engineer's  Unit  of  Mass — Summary. 


x  CONTENTS 

PAGE 

CHAPTER  XIX 

FORCE  WHICH  VARIES  AS  THE  DISPLACEMENT 301 

Force  of  a  Spring — Potential  Energy  of  a  Spring — Velocity  Pro- 
duced by  Elastic  Force — Vibration  from  Elastic  Force — Sudden 
Loads — Composition  of  Simple  Harmonic  Motions — Correction  for 
the  Mass  of  the  Spring — Determination  of  g — Determination  of  an 
Absolute  Unit  of  Force — Positive  Force  which  Varies  as  the 
Displacement — Summary. 

CHAPTER  XX 

CENTRAL  FORCES 327 

Definition — Work  of  a  Central  Force — Force  Inversely  as  the 
Square  of  the  Distance — Equipotential  Surfaces — Summary. 

CHAPTER  XXI 

ANGULAR  DISPLACEMENT  AND  VELOCITY 334 

Angular  Displacement — Angular  Velocity — Kinetic  Energy  of 
Rotation — Rotation  and  Translation — Translation  and  Rotation 
Reduced  to  Rotation — Summary. 

CHAPTER  XXII 

ACCELERATION  TOWARD  THE  CENTER 342 

Components  of  Acceleration — Acceleration  toward  the  Center — 
Centrifugal  Force  —  Static  Balance  —  Running  Balance — Ball 
Governor — Loaded  Governor  —  Fly-wheel  Stresses — Summary. 

CHAPTER  XXIII 

ANGULAR  ACCELERATION 354 

Angular  Acceleration — Displacement  with  Constant  Acceleration 
— Acceleration  and  Torque — Equivalent  Mass — Reactions  of 
Supports — Reaction  by  Experiment — Reactions  of  a  System — 
Summary. 

CHAPTER  XXIV 

ANGULAR  VIBRATION 368 

Work  of  Torque — Angular  Velocity  with  Variable  Torque — 
Time  of  Vibration — The  Gravity  Pendulum — Simple  Pendulum — 
Axis  of  Oscillation — Exchange  of  Axes — Summary. 

CHAPTER  XXV 

MOMENTUM  AND  IMPULSE 379 

Momentum  —  Impulse  —  Action  and  Reaction  —  Collision  of 
Inelastic  Bodies  —  Collision  of  Elastic  Bodies  —  Moment  of 
Momentum — Center  of  Percussion — Summary. 


CONTENTS  xi 

PAGE 

CHAPTER  XXVI 

ENERGY  TRANSFER 395 

Units  of  Energy — Power — Mechanical  Equivalent  of  Heat — 
Power  of  a  Jet  of  Water — Work  of  an  Engine — Friction  Brake — 
Cradle  Dynamometer — Transmission  Dynamometer — Power 
Transmission  by  Impact — Summary. 

INDEX.    .....  .    413 


NOTATION 

Symbols  frequently  used  in  this  book  are: 

a  =  linear    acceleration;    apparent    moment    arm;    length    of 

balance  beam;  radius  of  circle;  distance  on  figure, 
a  =  a  vector  of  length  a. 
A  =  Area;  a  force  (in  few  cases). 
b  =  breadth;  base  of  triangle;  a  distance. 
b  =  a  vector  of  length  b. 
B  =  a  force  (in  few  cases). 
c  =  a  distance;  a  constant  of  the  catenary;  distance  of  center 

of  gravity  of  balance  beam  below  central  knife-edge, 
c  =  a  vector  of  length  c. 

C  =  integration  constant;  a  force  (in  few  cases). 
d  =  diameter;  a  distance;  pitch  of  screw;  distance  of  central 
knife-edge    above   end   knife-edges;    distance    between 
parallel  axes, 
d  =  a  vector  of  length  d. 
e  =  base  of  natural  logarithms;  a  distance. 
E  =  electromotive  force;  modulus  of  elasticity. 
/  =  coefficient  of  friction. 
F  =  force;  total  force  of  friction. 
Fw  =  weight. 

g  =  acceleration  of  gravity;  a  constant,  32.174. 
h  =  height;  height  of  triangle. 
hp  =  horsepower. 
H  =  product  of  inertia. 

HQ  =  product  of  inertia  for  axes  through  center  of  gravity. 
H,  H,  HX)  H2  =  horizontal  force;  horizontal  vector. 
/  =  moment  of  inertia;  electric  current. 

/o  =  moment  of  inertia  for  axis  through  the  center  of  gravity. 
Ix  =  moment  of  inertia  with  respect  to  the  X  axis. 
/j,  =  moment  of  inertia  with  respect  to  the  Y  axis. 
I  max,  I min  =  maximum  and  minimum  moments  of  inertia. 
J  =  product  of  inertia. 
k  =  radius  of  gyration;  a  constant. 

fco  =  radius  of  gyration  for  axis  through  the  center  of  gravity. 
K  =  force  which  deforms  a  spring  1  foot;  a  constant;  integration 

constant;  coefficient  of  discharge. 
I  =  length;  length  of  simple  pendulum. 
m  =  mass  in  pounds,  grams,  or  kilograms. 
m.e.p.  =  mean  effective  pressure. 
M  =  moment. 

xiii 


xiv  NOTATION 

N  =  force  normal  to  surface. 
p  =  a  small  weight. 
P,  P  =  a  force;  force  which  causes  an  acceleration. 

Q  =  quantity  of  liquid. 
Q,  Q  =  a  force. 

r  =  radius;  amplitude  of  vibration. 
r  =  radius  to  center  of  gravity. 
R  =  resistance  in  ohms. 
R,  R  =  resultant  force;  reaction  of  support. 

s  =  length;  length  of  catenary  from  lowest  point. 
S,  S  =  equilibrant;  a  force;  unit  stress. 
t  =  time;  time  of  vibration;  thickness. 
tc  =  time  of  a  complete  period. 
T  =  torque. 

T,  T  =  tension;  tension  in  a  cord. 
U  =  kinetic  energy;  work. 
v  =  linear  velocity. 
V  =  volume. 
V,  V  =  a  vertical  force. 

w  =  weight  per  unit  length. 
w'  =  weight  per  unit  of  horizontal  distance. 
W  =  weight. 
x,  y,  z  =  distances. 

x,y,z  =  coordinates  of  center  of  gravity. 
X,  Y,  Z  =  coordinate  axes. 

yc  =  distance  to  center  of  pressure. 

a  =  angular  acceleration;  any  angle;  angle  with  X  axis;  angle 

of  contact  with  belt. 
0  =  angle  with  Y  axis;  any  angle. 
7  =  angle  with  z  axis;  any  angle. 
dx,  dy,  dz  =  small  increments  of  x,  y,  and  z. 
p  =  density. 

</>  =  an  angle;  angle  of  friction. 
6  =  an  angle;  angular  displacement. 
S  =  a  summation, 
w   =  angular  velocity. 


MECHANICS 

CHAPTER  I 
FUNDAMENTAL  IDEAS 

1.  Mechanics. — Mechanics  is  the  science  which  treats  of 
the  effect  of  forces  upon  the  form  or  motion  of  bodies.  The 
science  of  mechanics  is  divided  into  statics  and  dynamics. 
When  the  forces  which  act  on  a  body  are  so  balanced  as  to  cause 
no  change  in  its  motion,  the  problem  of  finding  the  relations  of 
these  forces  falls  under  the  division  of  statics.  When  the  forces 
which  act  on  a  body  cause  some  change  in  its  motion,  the  problem 
of  finding  the  relation  of  the  forces  to  the  mass  of  the  body 
and  to  the  change  of  its  motion  falls  under  the  division  of  dy- 
namics. The  division  of  dynamics  is  frequently  called  kinetics. 


2.  Illustrations. — Fig.  1  is  an  example  of  a  problem  of  statics. 
The  figure  shows  a  10-pound  mass  which  is  supported  by  two 
spring  balances.  In  Fig.  1,  I,  the  balances  are  nearly  vertical. 
One  balance  reads  a  little  less  than  5  pounds  and  the  other 
balance  reads  over  6  pounds.  In  Fig.  1,  II,  each  balance  makes 
a  large  angle  with  the  vertical.  The  right  balance,  which  is 
nearly  horizontal,  reads  13  pounds  and  the  left  balance  reads 
15  pounds.  In  this  position,  the  reading  of  each  balance  is 
greater  than  the  entire  weight  which  is  supported  by  the  two 
balances.  In  each  position,  the  balance  which  is  the  more  nearly 

1 


2 


MECHANICS 


[ART.  3 


vertical''  giv^s  -the  'larger  reading.  The  problem  of  finding  the 
relation  between  the  pulls  which  these  balances  exert  and  the 
angles  which  they  make  with  the  vertical  is  a  problem  of  statics. 

In  Fig.  2,  the  mass  of  10  pounds  is  supported  by  one  spring 
balance  and  by  a  cord  which  runs  over  a  pulley  and  carries  a 
mass  of  8  pounds  on  its  free  end.  This  system  will  come  to  rest 
in  a  definite  position.  If  moved  from  this  position,  the  system 
will  return  to  it  after  a  few  vibrations.  The  problem  of  finding 
this  position  and  the  tension  in  the  spring  balance  is  a  problem 
of  statics. 

If  the  cord  which  runs  over  the  pulley  of  Fig.  2  is  cut  or  broken, 
the  10-pound  mass  will  swing  back  and  forth  as  a  pendulum  and 
will  finally  come  to  rest  with  the  balance  in  a  vertical  position. 


FIG.  2. 


FIG.  3. 


This  final  position  is  shown  in  Fig.  3.  The  8-pound  mass  will 
fall  vertically  downward  with  increasing  speed.  As  the  10- 
pound  mass  swings  back  and  forth,  the  pull  on  the  balance  will 
change.  The  problem  of  finding  the  position  of  either  body  at 
any  time  after  the  cord  has  been  severed,  and  the  problem  of 
finding  the  pull  of  the  spring  balance  are  problems  of  dynamics. 

3.  Fundamental  Quantities. — The  problem  of  Fig.  2  involves 
position  and  direction.  These  are  properties  of  space.  The 
problem  also  involves  the  pull  of  the  cord  and  of  the  spring 
balance.  These  are  forces.  Every  problem  of  statics  involves 
the  two  fundamental  ideas  of  space  and  force.  If  a  body  of 
different  mass  were  used  in  place  of  the  10  pounds,  the  force 
and  space  relations  would  be  changed.  In  this  problem,  the 
forces  depend  upon  the  masses.  All  problems  of  statics  involve 
force  and  space.  Most  problems  of  statics  involve  also  mass. 

A  problem  of  dynamics  differs  from  one  of  statics  in  the  fact 
that  it  involves  the  element  of  time. 


CHAP.  I]  FUNDAMENTAL  IDEAS 

The  four  fundamental  quantities  of  mechanics  are  space, 
mass,  force,  and  time.  A  problem  of  dynamics  includes  all  four 
of  these  quantities.  A  problem  of  statics  may  include  all  except 
time. 

Space,  mass  (or  matter),  force,  and  time  are  elementary. 
None  of  them  may  be  reduced  to  anything  more  simple.  Con- 
sequently, it  is  useless  to  attempt  to  define  them.  On  the 
other  hand,  everyone  has  a  clear  knowledge  of  these  quantities. 
This  knowledge  has  been  gained  through  one  or  more  of  his 
senses  as  a  part  of  the  experience  of  his  lifetime. 

4.  Standards  and  Units. — While  space,  mass,  time,  and  force 
can  not  be  defined,  they  may  all  be  measured.     To  measure 
any  quantity,  it  is  necessary  to  have  a  unit  of  measure.     If 
measurements  are  to  be  taken  at  diverse  times  and  places,  it  is 
necessary  to  have  some  standard  unit  to  which  all  other  units  are 
referred.     These    units    of    measure  were  originally  arbitrarily 
chosen.     Other  units  might  have  been  selected  just  as  well. 
When  a  particular  unit  has  once  been  adopted,  however,  it  is 
important  that  its  value  be  preserved  without  change,  in  order 
that  physical  measurements  separated  by  wide  intervals  of  time 
may  be  accurately  compared. 

5.  Length. — Space  in  one  direction  is  length.     There  are  two 
official  standards  of  length  preserved  by  the  Bureau  of  Standards 
at  Washington.     These  are  the  Standard  Yard,  which  is  practi- 
cally equal  to  the  British  Imperial  Standard  Yard,  and  the  Stand- 
ard Meter,  which  is  a  copy  of  the  International  Standard. 

The  length  of  the  International  Meter  in  terms  of  the  wave 
length  of  cadmium  vapor  light  has  been  carefully  determined  by 
Michelson.  If  this  standard  bar  and  the  copies  preserved  by 
various  nations  should  undergo  any  change,  the  magnitude  of  the 
variation  may  be  found  by  a  new  comparison  with  the  wave 
length  of  this  light. 

The  foot  is  the  unit  of  length  which  is  commonly  used  by 
American  engineers.  A  foot  is  one-third  the  length  of  the 
Standard  Yard.  Physicists  use  the  centimeter  as  the  unit.  In 
countries  where  the  metric  system  has  been  adopted,  engineers 
employ  the  meter  as  the  unit  of  length. 

Length  measurements  are  generally  made  by  means  of  the 
sense  of  sight.  Sometimes  the  sense  of  touch,  the  sense  of  hear- 
ing, or  the  muscular  sense  is  used  in  comparing  two  lengths. 
The  vision,  however,  is  employed  to  get  the  actual  reading. 


4  MECHANICS  [ART.  6 

Space  in  one  dimension  is  length;  space  in  two  dimensions 
is  area;  space  in  three  dimensions  is  volume. 

The  idea  of  space,  including  length  and  direction,  is  gained 
by  the  child  through  the  sense  of  touch,  the  muscular  sense,  and 
the  sense  of  sight.  The  sense  of  hearing  also  assists  in  deter- 
mining direction. 

The  relation  between  the  metric  system  and  the  inch  is  given 
with  sufficient  accuracy  for  most  purposes  by 

39.370  inches  =  1  meter. 
2.540  centimeters  =  1  inch. 

Problems 

1.  Calculate  the  length  of  a  foot  in  centimeters  and  memorize  the  result. 

2.  Calculate  the  length  of  a  meter  in  feet  and  memorize  the  result. 

3.  Find  the  length  of  a  kilometer  in  feet  and  in  miles  and  compare  the 
results  with  some  reference  book. 

4.  Express  100  meters  in  yards  and  440  yards  in  meters. 

5.  By  logarithms  find  the  number  of  square  inches  and  square  feet  in 
one  square  meter. 

6.  A  hectare  is  100  meters  square.     Find  the  value  of  a  hectare  in  acres. 

7.  Using  five-place  logarithms,  find  the  number  of  cubic  centimeters  in 
one  cubic  inch.     Compare  with  some  handbook. 

8.  A  liter  is  a  cubic  decimeter.     Find  the  relation  between  the  liter  and 
the  U.  S.  liquid  quart. 

6.  Time. — The  standard  of  time  measurement  is  the  mean  solar 
day.     The  unit  commonly  employed  in  problems  of  mechanics  is 
the  mean  solar  second.     The  subdivision  of  the  solar  day  into 
hours,  minutes,  and  seconds  is  made  by  means  of  the  vibration 
of  pendulums  or  other  mechanical  devices.    In  making  time  meas- 
urements, the  senses  of  sight  and  hearing  are  used  in  connection 
with  these  mechanical  timepieces. 

The  child  gains  his  ideas  of  time  from  the  succession  of  events 
as  revealed  through  any  or  all  of  his  senses. 

7.  Matter  and  Mass. — From  the  mechanical  standpoint,  at 
least,  time  and  space  are  simple  and  easily  measured.     Time 
possesses  a  single  property,  that  of  extent;  space  has  extension 
or  length  in  three  dimensions.     Matter,  on  the  other  hand,  pos- 
sesses many  properties,  some  one  of  which  must  be  selected  and 
defined  as  measuring  the  amount  of  material  in  a  given  body. 
Volume  might  be  chosen  as  the  measure  of  the  quantity  of  mate- 
rial.    It  is  found,  however,  that  the  amount  of  a  given  material 
in  a  given  volume  may  be  greatly  changed  by  pressure.     It  is 


CHAP.  I]  FUNDAMENTAL  IDEAS  5 

also  found  that  equal  volumes  of  different  materials  differ  greatly 
in  their  mechanical  effects.  It  is  evident,  then,  that  some  other 
property  must  be  selected  to  designate  the  amount  of  matter. 

In  Fig.  4,  A  represents  a  block  of  soft  rubber  resting  on  some 
convenient  support.  In  Fig.  4,  II,  a  body  B  has  been  placed  on 
this  block  of  rubber.  The  length  of  the  block  is  found  to  have 
been  shortened.  If  B  is  removed,  the  block  A  returns  to  its 
original  length.  If  a  body  C  is  now  placed  on  A,  there  is  again 
a  change  in  length.  If  the  change  in  length  of  A  due  to  the  body 
C  is  the  same  as  that  due  to  the  body 
B,  the  bodies  B  and  C  are  said  to  con- 
tain equal  amounts  of  material.  The 
amount  of  material  (or  matter)  in  a 
body,  as  thus  defined  is  called  the  mass 
of  the  body.  Two  bodies  have  equal 
masses  if  they  produce  equal  deforma- 
tions in  a  third  body  when  they  are  applied  to  it  in  exactly  the 
same  way.  The  ordinary  spring  balance  is  a  common  form  of 
a  third  body  for  the  comparison  of  masses. 

Instead  of  being  supported  on  an  elastic  body,  the  bodies 
B  and  C  may  be  carried  on  the  hand  or  shoulder  of  the  observer. 
The  deformation  of  his  muscles  is  accompanied  by  a  sensation, 
called  the  muscular  sense,  which  enables  him  to  judge  roughly 
which  body  has  the  -greater  mass. 

A  second  method  of  measurement  of  mass  is  by  means  of 
inertia.  This  involves  the  conditions  of  change  of  motion  and 
will  be  considered  in  Chapter  XVII. 

The  child  gains  an  idea  of  mass  in  the  mechanical  way  by  means 
of  the  muscular  sense  and  the  sense  of  touch  as  experienced  when 
he  supports  bodies  free  from  the  earth,  stops  them  when  moving, 
or  otherwise  changes  their  motion.  The  concept  of  matter  in 
general  is  gained  through  all  the  senses. 

The  pound  is  the  common  unit  of  mass.  In  the  metric  system 
the  unit  is  the  kilogram.  Physicists  and  chemists  use  chiefly 
the  gram.  Units  of  mass  are  generally  called  "  weights."  A 
so-called  10-pound  weight  as  used  on  a  beam  balance  is  a  10- 
pound  mass. 

To  convert  from  the  metric  system  to  the  avoirdupois  system, 
the  relations  are, 

15432  grains  =  1  gram, 
453.6  grams    =  1  pound. 


6  MECHANICS  [ART.  8 

The  official  Standard  Pound  and  Kilogram  are  preserved  by 
the  Bureau  of  Standards: 

Problems 

1.  Find  the  number  of  pounds  in  1  kg.  correct  to  four  significant  figures. 

2.  The  mass  of  one  cubic  centimeter  of  water  at  4°C.  is  practically  one 
gram.     Find  the  mass  of  one  cubic  foot  of  water  in  pounds. 

8.  Force. — In  Fig.  4,  the  rubber  block  A  is  shortened  when  the 
body  B  is  placed  on  it.     The  body  B  is  said  to  exert  a  force  on 
A  at  their  surface  of  contact.     The  block  A    is   also   said   to 
exert  a  force  on  B  in  the  opposite  direction.     Force  is  some- 
thing which  may  exist   between  two  bodies  or  between   two 
parts  of  the  same  body.     Forces  occur  in  pairs.     There  is  a 
force  from  the  first  body  to  the  second  body,  and  an  equal  and 
opposite  force  from  the  second  body  to  the  first  body.     Newton 
stated  this  fact  in  what  is  called  the  Third  Law  of  Motion: 
"  Action  and  reaction  are  equal  and  opposed  to  each  other." 

A  force  always  causes  some  change  in  the  dimensions  of  a 
body.  A  force  always  tends  to  produce  some  change  in  the  motion 
of  the  body  upon  which  it  acts,  and  does  cause  some  change 
unless  it  is  balanced  by  an  equal  and  opposite  force  acting  on  the 
body,  or  by  a  number  of  forces  equivalent  to  an  equal  and  oppo- 
site force. 

Force  is  recognized  and  measured  by  means  of  the  change  in 
the  dimensions  and  form  of  elastic  bodies,  by  the  muscular  sense, 
and  by  the  change  in  the  motion  of  bodies  of  known  mass. 

9.  Weight. — When  the  masses  of  two  bodies  are  compared  by 
means  of  the  muscular  sensation  experienced  in  lifting  them,  the 
observer  really  gets  a  comparison  of  two  forces.     The  comparison 
of  mass  is  indirect.     If  the  observer  were  at  the  center  of  the 
Earth,  there  would  be  no  muscular  sensation  so  long  as  there  were 
no  change  in  the  motion  of  the  body.     The  Earth  exerts  a  force 
on  all  bodies.     This  force  is  in  the  form  of  a  pull  directed  toward 
the  center  of  the  Earth.     This  pull  or  attraction  is  called  the 
weight  of  the  body.     When  a  body  is  supported  and  thus  pre- 
vented from  moving  toward  the  earth,  the  support   exerts  a 
force  upward  which  is  equal  to  the  weight  of  the  body. 

When  a  physicist  speaks  of  the  weight  of  a  body,  he  always  means 
the  force  with  which  the  Earth  attracts  the  body.  In  the  com- 
mon use  of  the  word,  weight  generally  means  mass.  When  one 
says  that  the  weight  of  a  bar  of  iron  is  16  pounds,  he  is  usually 


CHAP.  I]  FUNDAMENTAL  IDEAS  7 

thinking  of  the  amount  of  iron  and  not  of  the  force  required  to 
lift  it.  Much  confusion  has  resulted  from  the  failure  to  designate 
clearly  which  of  these  two  meanings  is  intended. 

10.  .Relation  of  Mass  to  Weight. — The  definition  of  mass  in 
Art.  7  may  now  be  extended.  Two  bodies  have  equal  masses  if, 
at  a  given  point,  they  are  attracted  toward  the  Earth  with  equal  force. 
The  determination  of  mass  by  means  of  a  spring  balance  or  a 
beam  balance  is  accomplished  indirectly  by  a  comparison  of 
forces,  with  the  tacit  assumption  that  equal  forces  produce  equal 
effects.  The  first  definition  of  mass  is :  The  mass  of  a  body  is  pro- 
portional to  its  weight.  If  Fw  is  the  weight  of  the  body  in  some 
convenient  unit,  and  m  is  its  mass,  the  definition  may  be  ex- 
pressed algebraically  by  the  equation, 

Fw  =  km,  (1) 

in  which  k  is  a  constant.  The  numerical  value  of  k  depends  upon 
the  units  used  in  expressing  Fw  and  m.  These  units  may  be 
so  chosen  that  k  is  unity.  If  m  is  expressed  in  pounds  of  mass 
and  Fw  is  in  pounds  of  force,  then  k  =  1,  and 

Fw  =  m.  (2) 

Equation  (2)  states  that  the  mass  of  a  body  in  pounds  is  equal  to 
its  weight  in  pounds.1  The  word  pound  has  two  meanings  in 
mechanics.  It  may  be  used  to  designate  the  amount  of  material 
(mass)  or  to  express  the  force  of  attraction  toward  the  Earth 
(weight  as  meant  by  the  physicist).  In  a  similar  way,  the 
weight  of  one  kilogram  of  matter  is  one  kilogram,  and  the  weight 
of  one  gram  of  matter  is  one  gram.  With  the  systems  of  units 
in  everyday  use,  k  is  unity.  In  some  systems,  k  is  not  unity. 
In  the  absolute  system  of  units  k  =  g,  and  Fw  =  mg.  The  Weight 
of  a  mass  of  m  grams  in  that  system  is  mg  dynes. 

The  absolute  systems  of  units  are  not  used  by  engineers  in 
the  solution  of  problems  of  statics.  In  all  such  problems, 
Equation  (2)  applies.  The  weight  of  a  body  is  numerically 
equal  to  its  mass.  The  absolute  systems  of  units  and  a  second 

1  A  formula  is  merely  a  brief  statement  of  the  relation  of  quantities. 
The  letters  of  a  formula  represent  the  number  of  units  which  express  the 
magnitude  of  the  quantity.  In  the  above  equations,  m  represents  the  num- 
ber of  pounds,  the  number  of  kilograms,  the  number  of  tons,  or  the  number 
of  grams  of  material  in  the  body  under  consideration.  Similarly  Fw  repre- 
sents the  number  of  units  of  force  in  pounds,  kilograms,  tons,  grams,  poundals, 
or  dynes. 


8  MECHANICS  [ART.  11 

definition  of  mass  will  be  considered  in  this  book  in  Chapters 
XVII  and  XVIII. 

11.  Variation  of  Weight  with  Latitude  and  Altitude.— The 
weight  of  a  body  at  any  given  position  varies  as  its  mass.  It 
has  been  shown  by  experiment  that  the  weight  of  a  body  varies 
inversely  as  the  square  of  its  distance  from  the  center  of  the 
Earth.  If  the  Earth  .were  a  sphere  and  did  not  rotate  on  its 
axis,  the  weight  of  a  body  would  be  the  same  at  all  points  at  the 
same  level  on  its  surface.  Since  the  Earth  is  a  sphoroid  with  its 
polar  radius  about  13  miles  shorter  than  its  equatorial  radius, 
the  weight  of  a  body  increases  with  latitude.  While  a  pound 
mass  is  an  invariable  quantity,  the  weight  of  a  pound  mass,  as 
measured  by  a  spring  balance,  varies  with  the  latitude.  If  two 
masses  have  equal  weights  at  one  locality,  their  weights  will  be 
equal  at  any  other  locality.  Masses  may  be  compared  by  weigh- 
ing at  any  point.  A  spring  balance,  however,  which  has  been 
calibrated  by  means  of  a  standard  weight  at  one  locality,  can 
not  be  used  for  the  accurate  determination  of  mass  at  another 
locality.  (No  one  would  do  so  on  account  of  the  variation  of  the 
spring,  even  if  the  force  of  gravity  were  constant.) 

Since  the  practical  units  of  force  are  determined  from  the 
weights  of  the  standard  units  of  mass,  it  is  necessary  to  choose 
some  standard  location  for  the  definition  of  these  units  of  force. 
The  sea  level  at  45°  latitude  is  taken  as  this  standard  location. 

A  pound  force  is  defined  as  the  weight  of  a  pound  mass  at  the 
standard  location.  A  pound  mass  will  weigh  0.997  Ib.  at  the 
Equator  and  1.003  Ib.  at  the  Poles  on  a  spring  balance  which  is 
correct  at  the  standard  latitude.  This  difference,  while  impor- 
tant in  the  determination  of  physical  constants,  is  usually  neg- 
lected in  engineering  calculations. 

Problems 

1.  Taking  the  equatorial  diameter  of  the  Earth  as  8000  miles  and  the 
polar  diameter  as  26  miles  less,  and  neglecting  the  effect  of  the  rotation  of 
the  earth,  what  is  the  weight  at  the  Pole  of  a  body  which  weighs  1  pound 
at  the  Equator,  if  both  weighings  are  made  on  the  same  spring  balance? 

2.  What  is  the  relative  change  in  the  weight  of  a  body  when  it  is  taken 
from  a  point  at  the  sea  level  to  a  point  one  mile  higher? 


CHAPTER  II 
QUANTITY  AND  CALCULATIONS 

12.  Representation    of    Quantity    by    Numbers. — There    are 
several  ways  of  representing  the  magnitude  of  a  quantity.     The 
most  common  method  is  by  means  of  numbers,  as  6  feet,  8  pounds, 
10  seconds.     A  number  expresses  the  magnitude  of  the  quantity 
in  terms  of  the  unit  and  means  little  to  one  who  does  not  possess 
a  definite  idea  of  the  magnitude  of  the  unit.     Two  such  numbers 
give  a  clear  notion  of  the   relative  size  of  quantities    without 
conveying  any  information  as  to  the  actual  size  of  either.     Any- 
one  will   know   that   20   dekameters  is  twice    10   dekameters, 
without  having  any  idea  as  to  the  size  or  nature  of  a  dekameter. 
If  he  has  learned  that  a  'dekameter  is  10  meters  and  that  a  meter 
is  3.28  feet,  he  will  calculate  that  one  dekameter  is  approximately 
2  rods  and  that  20  dekameters  is  nearly  40  rods,  or  he  may  reduce 
to  yards  and  think  of  10  dekameters  as  a  little  over  100  yards. 
A  Frenchman,  on  the  other  hand,  who  thinks  in  the  metric  sys- 
tem, must  translate  rods  and  yards  into  meters  before  he  can 
have  a  real  idea  of  their  meaning. 

13.  Representation    of    Quantity    by    Lines. — The    relative 
magnitudes  of  several  quantities  are  frequently  represented  to 
the  eye  by  means  of  straight  lines  as  in  Fig.  5. 

Economic  data,  such  as  the  population  and  area 
of  countries  and  cities,  the  production  and  con- 
sumption of  commodities,  etc.,  are  commonly 
shown  in  this  way.  These  lines  may  be  hori- 
zontal, as  in  Fig.  5,  or  vertical  with  their  lower 
ends  on  the  same  horizontal  line. 

For  most  purposes  such  lines  are  merely  used 
to  express  magnitude  to  the  eye.     The  necessary 
calculations  are  made  by  means  of  numbers.     The  operation  of 
addition,   however,  may  be  performed  conveniently  with  lines. 
In  Fig.  6,  it  is  desired  to  find  the  sum  of  the  quantities  rep- 
resented by  the  lines  ab  and  cd.     The  lines  are  placed  together 
so  as  to  form  one  continuous  line  without  overlapping.     The  total 

9 


10  MECHANICS  [ART.  14 

line  thus  formed  is  the  sum  of  the  lines.  As  may  be  seen  from 
Fig.  6,  it  is  immaterial  in  what  order  the  lines  are  placed  together. 
For  subtraction,  especially  when  the  remainder  is  negative, 
it  is  desirable  to  adopt  some  convention  as  to  positive  and  nega- 
tive direction.  Horizontal  lines  extending  toward  the  right 
and  vertical  lines  extending  upward  are  regarded  as  positive. 
In  Fig.  6,  the  line  ab  runs  from  a  to  6.  The  left  end,  a,  is  called 
the  origin,  and  the  right  end,  b,  is  called  the  terminus.  To  find 
ab  +  cd,  the  origin  of  the  second  line  is  placed  at  the  terminus  of 


FIG.  7. 


3210 


01      234-567     09     IO 

7-3 
FIG.  8. 


a  b 

I         ab+cd 
I  a 


Ic         d 

cd  +  ab 

FIG.  6. 

the  first  line.  The  sum  of  the  two  lines  extends  from  the  origin 
of  the  first  line  to  the  terminus  of  the  second  line.  This  cor- 
responds with  ordinary  addition  of  numbers.  To  get  the  sum  of 
7  +  3  begin  at  7,  which  is  the  terminus  of  the  first  number,  and 
count  forward  3  steps.  This  is  shown  graphically  by  Fig.  7. 

Subtraction  is  the  addition  of  a  negative  quantity.  To  get 
7-3,  begin  at  7  and  count  backward  3  steps.  This  is  shown  by 
Fig.  8.  To  get  ab  —  cd,  begin  at  the  terminus  of  ab  and  measure 
the  length  of  cd  toward  the  left.  This  is  shown  in  Fig.  9.  The 
arrows  in  Fig.  9  give  the  direction  of  the  motion. 

14.  Vectors. — A  quantity  which  has  both  magnitude  and 
direction  is  called  a  vector  quantity.  Force  is  an  example  of 
this  kind  of  quantity.  In  Fig.  10,  ab,  cd,  and  e/are  vectors  in  the 
plane  of  the  paper.  The  vectors  ab  and  ef  are  equal,  since  they 
have  the  same  direction  and  equal  length. 


CHAP.  1IJ        QUANTITY  AND  CALCULATIONS  11 

In  the  vector  ab,  the  point  a  is  the  origin  and  the  point  b  is 
the  terminus.  The  vector  is  considered  as  extending  from  the 
origin  to  the  terminus,  as  is  indicated  by  the  arrow.  The  arrow 
points  from  the  origin  and  toward  the  terminus.  The  direction 
of  the  arrow  is  the  positive  direction  of  the  vector. 
When  a  vector  is  represented  by  a 

single    letter,    that    letter    is    usually     2 >        b 

printed   in   black   face  type.     In  Fig. 

11,  a  and  b  represent  two  such  vectors. 

The  arrow  shows  the  origin,  terminus,  e  f 

and  direction.  FIG        / 

A   vector  is  described  in  words  by 

giving  its  direction  and  its  length.  When  the  plane  of  the 
vector  is  known,  a  single  angle  is  sufficient  to  designate  its 
direction.  For  instance,  a  given  vector  is  8  feet  in  length  and 
makes  an  angle  of  35  degrees  with  the  horizontal.  When  the 
vectors  under  consideration  are  not  all  in  the  same  plane,  two 
angles  are  required  to  express  the  direction  of  each  vector.  For 
instance,  a  vector  is  8  feet  in  length  and  makes  an1  angle  of  40 
degrees  with  the  vertical  in  a  vertical  plane  which  is  north  25 
degrees  east. 

When  it  is  desirable  to  distinguish  a  quantity  which  has 
magnitude  but  not  direction  from  a  vector,  such  a  quantity  is 
called  a  scalar  quantity.  The  mass  of  a  body  or  the  number  of 
individuals  in  a  group  is  a  scalar  quantity.  In  an  algebraic 
formula  in  which  only  the  magnitude  of  a  vector  is  represented 
by  a  letter  while  its  direction  is  expressed  in  terms  of  angles,  the 

letter  is  printed  in  Italics  in- 
stead of  in  black  face  type. 
In  this  book,  a  letter  (such  as 
P  or  Q)  is  used  to  represent 
a  force,  which  is  a  vector. 
D  When  emphasis  is  put  on 

both  the  direction  and  mag- 
nitude of  the  force,  the  letter  is  printed  in  black  face  type.  When 
only  the  magnitude  is  stressed,  it  is  printed  in  Italics. 

15.  Addition  of  Vectors. — The  addition  of  vectors  is  denned 
in  the  same  way  as  the  addition  of  lines  (Art  13).  The  origin  of 
the  second  vector  is  placed  at  the  terminus  of  the  first  vector. 
The  line  which  extends  from  the  origin  of  the  first  vector  to 
the  terminus  of  the  second  vector  is  the  sum  of  the  two  vectors. 


12  MECHANICS  [ART.  16 

Fig.  11,  II,  shows  the  addition  of  the  vectors  a  and  b  to  get  their 
sum  a  +  b.  Fig,  11,  III,  shows  b  +  a.  The  vector  a  is  first 
in  Fig.  11,  II,  and  the  vector  b  is  first  in  Fig.  11,  III.  Fig.  11, 
IV,  shows  both  additions  in  one  diagram.  The  additions  begin 
at  a  common  origin  O.  Since  the  two  lines  a  are  equal  and  parallel 
and  the  two  lines  b  are  also  equal  and  parallel,  the  four  lines 
form  a  parallelogram.  The  diagonal  of  this  parallelogram  is 
the  vector  sum  and  it  is  immaterial  in  what  order  the  two  vectors 
are  added.  The  sum  of  three  vectors  is 
found  in  the  same  way.  Fig.  12,  II,  repre- 
sents the  sum  of  three  vectors.  Fig.  12, 
I,  may  be  regarded  as  a  graphical  statement 
°^  ^e  vec^ors  wmcn  are  to  be  added.  In 
^is  figure  all  the  vectors  start  from  a  com- 
mon  origin.  In  Fig.  11,  I,  on  the  other 
FIG.  12.  hand,  the  vectors  a  and  b  start  from  differ- 

ent origins. 

There  are  two  methods  of  finding  the  vector  sum.  These 
are  the  graphical  method  in  which  the  lengths  and  angles  are 
measured,  and  the  trigonometric  (or  algebraic)  method  in  which 
the  lengths  and  angles  are  computed. 


Problems 

1.  Given  two  vectors,  a  =  15  ft.  at  0  degrees,  b  =  12  ft.  at  40  degrees. 
Solve  graphically  for  the  vector  sum,  a  +  b.     Use  the  scale  of  1  inch  =  5 
feet.     Measure  the  vector  sum  and  express  the  result  in  feet.     Measure 
the  angle  of  the  vector  sum  with  the  first  vector  and  express  the  result  in 
degrees. 

First  construct  the  statement  to  scale  as  shown  in  Fig.  13,  I.  Then  draw 
a  in  Fig.  13,  II,  equal  and  parallel  to  a  of  the  statement.  From  the  terminus 
of  a  draw  b  equal  and  parallel  to  b  of  the  statement. 

2.  Solve  problem  1  for  the  vector  sum  b  -f-  a. 

3.  Find  the  sum  of  three  vectors:  20  ft.  at  0  degrees,  15  ft.  at  45  degrees, 
and  10  ft.  at  110  degrees.     Use  the  same  scale  as  in  Problem  1. 

4.  Find  the  vector  sum  of  16  ft.  at  10  degrees  and  20  ft.  at  70  degrees. 
Construct  the  10-degree  angle  by  means  of  its  tangent.     Construct  the 
60-degree  angle  by  means  of  its  chord.     Measure  the  angle  of  the  vector 
sum  by  means  of  its  chord  and  check  by  means  of  the  sine  of  the  angle  which 
the  vector  sum  makes  with  a  line  at  90  degrees. 

16.  Components  of  a  Vector. — The  sum  of  two  or  more  vectors 
is  frequently  called  the  resultant  vector  and  the  vectors  which 


CHAP.  II]        QUANTITY  AND  CALCULATIONS 


13 


are  added  are  called  components  of  the  resultant.  The  process 
of  finding  the  resultant  is  called  composition  of  vectors.  The 
process  of  finding  the  components  is 
called  resolution.  In  Fig.  13,  the  vector 
a  +  b  is  the  resultant  of  vectors  a  and 
b,  and  the  vectors  a  and  b  are  com- 
ponents of  a  +  b. 

Problems 


FIG.  13. 


1.  A  vector  of  25  ft.  at  30  degrees  is  made  up  of  two  components.     One 
of  these  components  makes  an  angle  of  5  degrees  with  the  reference  line 
and  the  other  makes  an  angle  of  45  degrees  with  the  reference  line.     Find 
these  components  graphically. 

2.  A  vector  of  20  ft.  at  45  degrees  is  made  up  of  a  vector  a  at  20  degrees 
and  a  vector  b  which  is  12  ft.  in  length.     Find  the  magnitude  of  a  and  the 
direction  of  b. 

The  most  important  kind  of  resolution 
of  vectors  is  that  in  which  each  com- 
ponent is  formed  by  the  orthographic 
projection  of  the  resultant  upon  a  line 
along  the  desired  direction.  In  Fig.  14, 
c  is  a  vector  which  makes  an  angle  6 
with  the  horizontal.  Its  horizontal 
component  is  a  and  its  vertical  com- 


FlG. 


ponent  is  b.     The 
the  equations 


lengths   of  these   components  are  given  by 


a  =  'c  cos  0. 


b  =  c  sin  0. 


(In  these  equations,  the  letters  a,  6,  and  c  represent  magnitudes 
only.  For  that  reason  they  are  printed  in  Italics  instead  of  in 
black  face  type.) 

Figure   15  shows  two  such  orthographic  components,  which 
with  their  resultant  form  a  right  triangle. 

When  the  term  component  is  used  without 
qualification,  the  orthographic  component  is 
generally  meant.  The  process  of  finding  the 
orthographic  component  of  a  vector  in  a  given 
direction  is  called  resolution  in  that  direction. 

Problems 

Solve  the  following  problems  graphically  and  trigonometrically. 
3.  A  vector  of  60  ft.  is  north  27  degrees  east.     Find  its  component  north 
and  its  component  east.  Ans.     53.46  ft.  north;  27.24  ft.  east. 


14  MECHANICS  [ART.  17 

4.  A  vector  of  45.67  ft.  is  directed  north  27°  07'  west.     Find  its  component 
north  and  its  component  west  by  means  of  logarithmic  functions. 

5.  Find  the  component  of  the  vector  of  Problem  3  along  a  line  which  is 
north  45  degrees  east.     Check  by  means  of  the  sum  of  the  components 
along  this  direction  of  the  east  and  north  components  as  given  in  the  answer 
to  Problem  3. 

6.  The  horizontal  component  of  a  vector  is  18.24  ft.  and  the  vertical 
component  is  12.48  ft.     By  means  of  the  tangent  find  the  angle  which  the 

vector  makes  with  the  horizontal.  Then 
find  the  length  of  the  resultant  vector  by 
means  of  the  cosine  (or  secant)  of  this  angle. 
Check  by  projecting  the  horizontal,  and  ver- 
tical components  upon  the  line  of  the  re- 
sultant and  adding  the  components  thus 
found  (Fig.  16). 

7.  The  horizontal  component  of  a  vector 
is    27.734   ft.     The    vertical  component  is 
16  18.245ft.    Find  the  direction  of  the  resultant 

vector  by  means  of  the  logarithmic  tangent. 

Find  the  magnitude  of  the  resultant  by  means  of  the  logarithmic  cosine. 

Check  by  projections  as  in  Problem  16. 

17.  Computation  of  the  Vector  Sum. — Problem  6  of  the  pre- 
ceding article  is  an  example  of  the  computation  of  the  vector  sum 
when  there  are  only  two  vectors  to  be  added,  and  these  vectors 
are  at  right  angles  to  each  other.  In  Fig.  13,  only  two  vectors 
are  to  be  added,  but  these  are  not  at  right  angles  to  each  other. 
In  this  problem,  the  vector  sum  may  be  found  by  the  formulas 
for  oblique-angled  triangles.  The  unknown  side  may  be  com- 
puted by  the  Law  of  Cosines  and  the  angles  then  found  by  the 
Law  of  Sines.  Another  method  is  to  find  first  the  unknown 
angles  by  means  of  the  formula  which  expresses  the  relation  of  the 
tangents  of  half  the  sum  and  half  the  difference  of  two  angles  to 
the  sum  -and  difference  of  the  sides  opposite,  and  then  to  find 
the  unknown  side  by  the  Law  of  Sines.  This  last  method  permits 
the  use  of  logarithmic  functions.  Since  neither  of  these  methods 
is  convenient  to  apply  when  there  are  more  than  two  vectors 
to  be  added,  it  is  best  to  learn  a  general  method  which  is  valid 
for  all  cases  of  vectors  in  a  single  plane. 

First,  find  the  component  of  each  vector  in  the  direction  of 
some  line  in  their  plane.  The  algebraic  sum  of  these  components 
is  the  component  of  the  vector  sum  in  this  direction.  Then  find 
the  component  of  each  vector  in  the  direction  of  a  second  line 
which  is  at  right  angles  to  the  first  direction.  The  algebraic 


CHAP.  II]        QUANTITY  AND  CALCULATIONS 


15 


sum  of  these  components  is  the  component  of  the  vector  sum 
in  this  direction.  These  two  algebraic  sums  represent  the  legs 
of  a  right-angled  triangle.  The  hypotenuse  of  this  triangle 
is  the  vector  sum  required. 

Figure  17  shows  the  addition  of  three  vectors  a,  b,  and  c  in  the 
same  plane.  These  vectors  make  angles  of  a,  fi,  and  7,  respec- 
tively with  the  horizontal  line.  If  H  is  the  algebraic  sum  of  the 
components  of  these  vectors  parallel  to  the  horizontal  line, 

H  =  a  cos  a.  +  b  cos  ff  -f  c  cos  7. 
If  V  is  the  algebraic  sum  of  all  the  vertical  components, 

V  =  a  sin  a  +  b  sin  0  +  c  sin  7. 


FIG.  17. 

The  resultant  vector  is  the  vector  sum  of  a  horizontal  vector  of 

length  H  and  a  vertical  vector  of  length  V.     (Fig.  17,  III.) 

If  B  is  the  angle  which  the  resultant  vector,  makes  with  the 
horizontal, 

tan  B    =  ^;  (1) 

*~  (2) 


R  =  H  seed   = 


cosO 


R    = 


sin  B' 


(3) 


If  H  is  greater  than  F,  use  Equation  (2)  to  find  R;  if  V  is  greater 
than  H,  use  Equation  (3).  The  most  accurate  result  is  obtained 
by  this  method. 

It  is  not  necessary  that  the  direction  of  H  should  be  horizontal. 
The  resolution  may  be  taken  along  any  direction,  and  the  resolu- 
tion for  V  at  right  angles  to  the  direction  of  H. 


16 


MECHANICS 


[ABT.  17 


Example 

Find  the  direction  and  magnitude  of  the  vector  sum  of  12  units  at  18 
degrees  and  15  units  at  50  degrees. 

The  equation  for  the  components  at  0  degrees  is, 

H  =  12  cos  18°  +  15  cos  50°, 

12  X  0.9511  =  11.4132 

15  X  0.6428  =  9.6420 

H  =  21.0552 

The  equation  for  the  components  at  90  degrees  is, 

V  =  12  sin  18°  +  15  sin  50°, 
12  X  0.3090  =    3.7080 
15  X  0.7660  =  11.4900 


Tan,  = 


R 


21.0552 
35°  49'. 
21.0552 

cos  e 


V  =  15.1980 
=  0.7218, 


21,0552 
0.810S 


FIG.  18. 

Instead  of  taking  resolutions  along  the  lines  at  0  degrees  and  90  degrees, 
resolve  along  the  direction  of  the  first  vector  and  perpendicular  to  that 
direction  (Fig.  18).  If  the  component  of  the  vector  sum  along  the  first 
direction  is  H'  and  the  component  at  right  angles  to  this  direction  is  V, 

H'  =  12  +  15  cos  32°, 

H'  =  12  +  15  X  0.8480  =  12  +  12.7200  =  24.7200, 

V  =  15  sin  32°  =  15  X  0.5299        =  7.9485. 


Tan  6 


24.7200 
0  =  17°  49'. 
R  =  24.7200 

cos  e 


=  0.3215, 


24.7200 


25.96  units. 


0.9520 

This  will  be  recognized  as  the  method  of  computing  oblique-angled  triangles 
by  means  of  right  triangles. 

The  computation  of  the  components  in  the  foregoing  ex- 
ample has  been  carried  to  four  decimal  places,  which  gives 
five  or  six  significant  figures  in  the  resultant  products.  As 


CHAP.  II]        QUANTITY  AND  CALCULATIONS 


17 


only  four-place  trigonometric  functions  have  been  used,  the 
last  figure  or  the  last  two  figures  may  be  incorrect.  The  last 
figure  of  each  product  may  well  be  dropped  and  the  results 
written  #  =  11.413  +  9.642  =  21.055.  Many  computers  would 
drop  the  last  two  figures  and  give  H  =  21.06.  This,  however, 
should  not  be  done.  An  error  of  unity  in  the  last  figure  of  21.06 
is  greater  relatively  than  an  error  of  unity  in  the  fourth  place 
of  the  cosines  which  are  used  in  the  computation.  For  accurate 
work  it  is  a  good  rule  to  carry  the  calculations  one  figure  farther 
than  the  data,  and  finally  drop  the  last  figure  from  the  result. 
This  method  prevents  errors  in  the  calculations,  which  may 
greatly  exceed  the  errors  of  the  data. 

In  the  foregoing  example,  it  has  been  assumed  that  the  length 
of  each  vector  is  correct  to  four  significant  figures.  In  many 
cases  the  degree  of  accuracy  is  expressed  by  the  addition  of 
zeros.  A  length  of  12  feet,  which  is  correct  to  hundredths  of  a 
foot,  is  written  12.00  ft.  This,  however,  is  not  always  done. 

In  these  .examples,  it  is  also  assumed  that  the  angles  are 
correct  /to  minutes.  An  angle  which  is  given  as  32°  will  be 
understood  to  be  32°  00'. 


Problems 

1.  Find  the  magnitude  and  direction  of  the  resultant  of  24.8  ft.  at  0 
degrees  and  22.8  ft.  at  65  degrees.  Ans.     40.16  ft.  at  30°  58'. 


30.8 


^ 

1 


FIG.   19. 


2.  Find  the  magnitude  and  direction  of  the  resultant  of  24  units  at  10 
degrees,  32  units  at  28  degrees,  and  16  units  at  70  degrees.     Resolve  along 
0  degrees  and  90  degrees. 

3.  Check  Problem  2  by  resolving  along  the  line  at  10  degrees  and  along 
the  line  at  100  degrees. 

2 


18  MECHANICS  [ART.  18 

4.  Find  the  direction  and  magnitude  of  the  vector  sum  of  25.6  units  at 
12  degrees,  18.3  units  at  56  degrees,  30.8  units  at  98  degrees,  and  21.4  units 
at  123  degrees. 

Where  there  are  several  vectors,  it  is  convenient  to  arrange  the  solution 
in  tabular  form. 


Length 

Angle 

Cosine 

Sine 

H  comp. 

T7  comp. 

25.6 

12° 

'    0.9781 

0.2079 

25  .  039 

5.322 

18.3 

56° 

0.5592 

0.8290 

10.233 

15.171 

30.8 

98° 

-0.1392 

0.9903 

-  4.287 

30.501 

21.4 

123° 

-0.5446 

0.8387 

-11.654 

17.948 

19.331          68.942 


6  =  74°  20'. 

68.942      68.942 


09628 


5.  Solve  Problem  4  by  resolutions  along  12  degrees  and  102  degrees. 

Ans.     H'  =  33.243;  V  =  63.419;  cotan  8  =  0.5242. 

6.  Solve  problem  1  for  the  vector  sum  by  means  of  the  law  of  cosines. 
Then  find  the  direction  by  the  law  of  sines. 

7.  Find  the  vector  sum  of  24.62  units  at  0  degrees  and  18.28  units  at  62 
degrees.     First,  find  the  angles  by  means  of  the  relations  of  the  tangents 
of  half  the  sum  and  half  the  difference.     Then  find  the  magnitude  of  the 
vector  sum  by  the  law  of  sines. 

8.  A  vector  of  24.6  units  at  20  degrees  and  a  vector  of  16.8  units  at  an 
unknown  angle  have  a  resultant  of  12.4  units.     Find  the  unknown  directions 
by  means  of  the  tangents  of  the  half  angles. 

18.  Vector  Difference.  —  Subtraction  is  the  addition  of  a 
negative  quantity.  If  a  negative  quantity  be  regarded  as 
having  direction,  its  direction  is  opposite  to  the  direction  of  a 
positive  quantity.  A  negative  vector  is  opposite  in  direction  to  a 
positive  vector.  If  vector  a  =  4  feet  at  20  degrees,  —  a  =  4 
feet  at  200  degrees. 

In  Fig.  8,  the  number  3  is  subtracted  from  7  by  counting 
backwards  three  units  from  the  terminus  of  7.  Ordinary  addi- 
tion and  subtraction  may  be  regarded  as  special  cases  of  vector 
addition  and  subtraction  in  which  all  the  positive  vectors  are  in 
the  same  direction.  Subtracting  3  is  equivalent  to  adding  —  3 
or  counting  three  units  in  the  negative  direction. 

Figure  20  shows  two  vectors,  aandb.  To  get  a  —  b,  the  vector 
a  is  first  drawn.  From  the  terminus  of  vector  a  of  Fig.  20,  II, 
the  vector—  b  is  drawn  opposite  .to  the  direction  of  vector  b  of 


CHAP.  II]        QUANTITY  AND  CALCULATIONS 


19 


Fig.  20, 1.  The  vector  c  from  the  origin  of  a  to  the  terminus  of 
— b  is  the  required  difference. 

a  —  b  =  c, 
a  =  c  +  b. 

Figure  20,  III,  shows  another  method  of  getting  the  same  result. 
If  c  =  a  —  b,  a  =  b  +  c.  The  vector  c  is  the  vector  which 
must  be  added  to  vector  b  to  get  vector  a.  Vectors  a  and  b 
are  drawn  from  the  same  origin  and  the  terminus  of  b  is  joined  to 
the  terminus  of  a.  This  vector  c,  with  its  origin  at  the  terminus 
of  b  and  its  terminus  at  the  terminus  of  a,  is  the  required  differ- 
ence. Fig.  20,  IV,  shows  the  two  methods  combined.  The 


is: 


vectors  a  and  b  form  the  sides  of  a  parallelogram.  The  diagonal 
which  starts  at  the  origin  of  the  first  two  vectors  is  the  vector 
sum.  The  other  diagonal  is  the  vector  difference. 


Problems 

1.  Given  a  vector  a  =  22  units  at  0  degrees  and  a  vector  b  =  22  units 
at  30  degrees.     Find  b  —  a.  Ans.     b  —  a  =  11.39  units  at  105°. 

2.  A  vector  a  =  17.28  ft.  at  56°  and  a  vector  b  =  27.34  ft.  at  24  degrees. 
Find  a  —  b.     Solve  graphically,  then  solve  trigonometrically. 

3.  A  vector  a  =  12.4  ft.  at  20°,  a  vector  b  =  17.2  ft.  at  45°,  and  a  vector 
c  =  19.2  ft.    at   64°.     Find   a  +  b  -  c.     First   solve   graphically.     Then 
solve  by  resolutions  as  in  Art.  17.  Ans.     15.42  ft.  at  356°  49'. 

4.  Check  Problem  3  by  resolutions  along  20°  and  110°. 

5.  In  Problem  3,  find  a  —  b  —  c. 

19.  Vectors  in  Space. — A  vector  in  a  plane  may  be  specified 
numerically  by  two  quantities.  These  may  be  its  length  and 
its  angle  with  some  axis,  which  correspond  with  polar  coor- 
dinates, or  its  components  along  two  axes,  which  correspond 
with  Cartesian  coordinates. 


20 


MECHANICS 


[AET.  19 


To  specify  fully  a  vector  in  space  requires  three  quantities. 
These  may  be  its  length  and  two  angles  or  its  components  along 
three  axes  which  are  not  all  in  one  plane. 

Figure  21  shows  one  way  cf  expressing  the  angles  of  a  vector  in 
space.  The  vector  of  length  /  makes  an  angle  0  with  the  Y 
axis,  while  the  plane  which  passes  through  the  vector  and  the 
Y  axis  makes  an  angle  <£  with  the  XY  plane.  This  is  equivalent 
to  co-latitude  and  longitude  on  a  sphere.  The  Y  axis  may  be 
regarded  as  the  polar  axis  of  the  sphere.  The  angle  Q  is  equivalent 
to  the  co-latitude  and  the  angle  $  is  equiva'ent  to  the  longitude. 
These  coordinates  are  called  spherical  coordinates. 


FIG.  21. 


FIG.  22. 


If  H  is  the  component  in  the  XZ  plane  and  V  is  the  component 
parallel  to  the  Y  axis,  Fig.  22, 

B  =  I  sin  0,  (1) 

V  =  I  cos  0.  (2) 

If  H x  is  the  component  along  the  X  axis  and  Hz  is  the  component 
along  the  Z  axis, 


Hx  =  H  cos  <£  =  /  sin  0  cos  <f>, 
Hz  =  H  sin  3>  =  /  sin  0  sin  <i>. 


(3) 
(4) 


Problems 


1.  A  vector,  25  feet  in  length,  makes  an  angle  of  35  degrees  with  the 
horizontal  in  a  vertical  plane  which  is  south  25  degrees  east.  Find  its 
vertical  component  and  its  horizontal  components  east  and  south. 

Ans.     V  -  14.34  ft.;  He  =  8.65  ft.;  tf,  =  18.56  ft. 


CHAP.  II]        QUANTITY  AND  CALCULATIONS 


21 


2.  A  vector,  64.24  feet  in  length,  is  in  a  vertical  plane  which  is  north 
37  degrees  east.     The  elevation  of  the  vector  is  47  degrees.     Find  its  com- 
ponent north,  its  component  east,  and  its  component  vertical. 

3.  The  vertical  component  of  a  given  vector  is  25.6  feet.     The  horizontal 
component  east  is  16.8  feet.     The  horizontal  component  south  is  14.4  feet. 
Find  the  direction  and  magnitude  of  the  vector. 

20.  Vectors  by  Direction  Cosines. — A  second  method  of 
expressing  the  direction  of  a  vector  is  by  means  of  the  angles 
which  it  makes  with  two  of  the 
coordinate  axes.  In  Fig.  23,  the 
vector  OP  of  length  I  is  drawn  as 
the  diagonal  of  a  rectangular 
parallelepiped.  Three  edges  of 
this  parallelepiped  lie  in  the  axes 
of  coordinates.  The  angle  be- 
tween the  vector  and  the  X  axis 
is  a.  The  angle  between  the  vector 
and  the  Y  axis  is  /?;  and  the  angle 
between  the  vector  and  the  Z  axis 
is  7.  These  angles  are  the  direction 
angles  of  the  vector,  and  their  cosines 
are  called  the  direction  cosines. 


FIG.  23. 


Hx  =  I  cos  a;  (1) 

V   =  I  cos  0;  (2) 

Hz  =  I  cos  7.  (3) 
Since  Z2  =  Hi  +  T2  +  H\  =  Z2(cos2a  +  cos20  +  cos27), 

cos2a  +  cos2j3  +  cos27  =  1  (4) 

Equation  (4)  is  a  fundamental  formula  of  solid  analytic  geometry. 

In  the  statement  of  a  problem,  two  of  the  direction  angles 
are  given  and  the  third  is  found  by  means  of  Equation  (4). 

It  will  be  noted  that  Equation  (2)  is  the  same  as  Equation  (2) 
of  Art.  19.  One  angle  is  measured  in  the  same  way  in  both 
methods. 

(In  works  of  mathematical  physics,  the  components  which  are 
here  written  Hx,  V,  and  Hz  are  written  X,  Y,  and  Z.) 

It  is  not  necessary,  of  course,  that  these  axes  be  horizontal 
and  vertical.  They  may  be  taken  in  any  convenient  direction. 
In  most  cases  of  practice,  however,  it  is  easiest  to  measure  the 
angles  from  a  vertical  line  and  from  lines  in  the  horizontal 
plane. 


22  MECHANICS  [ART.  21 

Example 

A  vector,  30  feet  in  length,  makes  an  angle  of  70  degrees  with  the  hori- 
zontal axis  east,  and  an  angle  of  65  degrees  with  the  vertical  axis.  Find 
its  components  along  each  of  the  coordinate  axis. 

cos27  =  1  -  cos270°  -  cos*  65°  =  1  -  0.1170  -  0.1786  =  0.7044; 

cosV-  0.7044  -1+%*2*; 

cos  2y  =  0.4088; 

2-y  =  65°52',  7  =  32°56'. 

The  squares  of  the  cosines  of  70°  and  of  65°  are  most  easily  found  by 
means  of  the  cosines  of  the  double  angles. 

Problems 

1.  A  vector  24  feet  in  length  makes  an  angle  of  65  degrees  with  the  south 
horizontal  line  and  an  angle  of  60  degrees  with  the  east  horizontal  line.  The 
direction  of  the  vector  is  above  the  horizontal.     Find  its  angle  with  the 
vertical  and  find  its  components. 

Ans.  ft  =  60°  00';  Hs  =  16.97  ft.;  He  =  12  ft.;  V  =  12  ft. 

2.  A  vector  40  feet  in  length  makes  an  angle  of  67  degrees  with  the  vertical 
and  an  angle  of  23  degrees  with  the  east  horizontal  axis.     Find  its  angle 
with  the  south  horizontal  axis  and  find  its  components. 

3.  A  vector  50  feet  in  length  makes  an  angle  of  25  degrees  with  the  east 
horizontal  axis  and  an  angle  of  80  degrees  with  the  vertical.     Its  direction 
is  east  of  south.     Find  its  angle  with  the  south  horizontal  axis  and  find  its 
components.     Ans.  7=  67°  20';  He  =  45.32ft.;  Ha  =  19.27ft.;  V  =  8.68ft. 

4.  The  east  component  of  a  vector  is  18  feet.     The  vertical  component 
is  20  feet.     The  length  of  the  vector  is  32  feet.     Find  the  south  component 
and  the  direction  angles. 

21.  Addition  of  Vectors  in  Space. — To  compute  the  vector 
sum  of  vectors  in  one  plane,  each  vector  is  revolved  along  two  axes 
at  right  angles  to  each  other.  The  sum  of  the  components  along 
one  axis  forms  one  leg  of  a  right-angled  triangle,  and  the  sum  of 
the  components  along  the  other  axis  forms  the  other  leg  of  this 
triangle.  The  hypotenuse  of  this  triangle  is  the  vector  sum  or 
resultant  of  the  separate  vectors.  (Art.  17.)  Likewise,  to  find 
the  vector  sum  of  a  number  of  vectors  in  space,  each  vector  is 
resolved  into  three  components  along  axes  which  are  mutually  at 
right  angles.  The  sum  of  the  components  along  each  axis  forms 
one  edge  of  a  rectangular  parallelepiped .  The  diagonal  of  this 
parallelepiped  is  the  vector  sum  required. 

Example 

Find  the  vector  sum  of  a  vector  20  feet  in  length,  which  is  elevated  40 
degrees  above  the  horizontal  in  a  vertical  plane  north  25  degrees  east,  and 


CHAP.  II]        QUANTITY  AND  CALCULATIONS  23 

a  vector  30  feet  in  length,  which  is  elevated  65  degrees  in  a  vertical  plane 
north  55  degrees  east. 

Vector.  V  Hn 

20         12.856         13.883 
30        27.189         7.272 


40.045         21.155         16.861 

In  horizontal  plane 

_  He  _   16.861 
tan0  -  ^  -  21155; 

log  tan  <f>  =  9.90147;  <f>  =  38°  33'. 

=  2JL155          ff  =  t  43215   H  =  27  05  ft 
cos  <t> 

Tan  0  =  |^^;  log  tan  ft  =  9.82960;  0  =  34°  02'. 
R  =  ^^j.iog  R  =  1.68415;  72  =  48.32  ft. 

Problems 

1.  Find  the  vector  sum  of  a  vector  30  feet  in  length,  which  is  elevated 
60  degrees  in  a  vertical  plane  north  35  degrees  east;  a  vector  20  feet  in  length, 
north  65  degrees  east  in  a  horizontal  plane;  and  a  vector  25  feet  in  length  in 
a  north  and  south  vertical  plane  at  an  angle  of  36  degrees  north  of  the 
vertical. 

Ans.  V  =  46.21  ft.;  Hn  =  35.43  ft.;  He  =  26.73  ft.;  0  =  37°  02';  H  = 
44.38  ft.;  0  =  43°  51;  R  =  64.08  ft. 

2.  In  Problem  1,  find  the  angle  which  the  resultant  makes  with  each  of 
the  three  coordinate  axes. 

3.  Find  the  vector  sum  of  the  following  vectors:  a  vector  20.66^ feet  in 
length,  which  is  elevated  32  degrees  in  a  vertical  plane  north  16  degrees 
east;  a  vector  12.84  feet  in  length,  which  is  elevated  64  degrees  in  a  vertical 
plane  north  30  degrees  east;  and  a  vector  18.62  feet  in  length,  which  is 
elevated  40  degrees  in  a  vertical  plane  north  45  degrees  west. 

Ans.  V  =  34.46  ft.;  Hn  =  31.80  feet.;  He  =  -  2.44  ft.;  R  =  46.95  ft. 
at  an  angle  of  42°  47'  with  the  vertical  in  a  plane  north  4°  23'  west. 

4.  Find  the  vector  sum  of  a  vector  25  feet  in  length  in  the  north  east 
quadrant  at  an  angle  of  64  degrees  with  the  vertical  and  at  an  angle  of  60 
degrees  with  the  north  horizontal  axis;  and  a  vector  20  feet  in  length  at  an 
angle  of  40  degrees  with  the  vertical  in  a  vertical  plane,  which  is  north  36 
degrees  east. 

22.  Graphical  Resolution  of  Vectors  in  Space. — Fig.  24  shows 
the  graphical  method  of  finding  the  components  of  a  vector 
when  the  directions  are  given  in  spherical  coordinates.  The 
vector  of  length  I  makes  an  angle  ft  with  the  vertical  in  a  vertical 


24 


MECHANICS 


[ART.  22 


VP/ane 


HP/ane 


plane  at  an  angle  0  with  the  XY  vertical  plane.  The  lineOP,of 
length  I,  is  first  drawn  in  the  7  plane  at  an  angle  0  with  the  verti- 
cal. Its  projection  on  the  Y  axis  gives 
the  V  component,  and  its  projection  on 
the  X  axis  gives  the  length  of  the  hori- 
zontal component  H.  The  line  OQ  of 
length  H  is  revolved  through  an  angle  <f>  in 
the  horizontal  plane  to  the  position  OQ' . 
This  line  OQ'  gives  the  location  and  mag- 
nitude of  the  horizontal  component.  The 
component  of  OQ'  along  the  X  axis  is  Hx 
and  the  component  along  the  Z  axis  is 
Hz. 

Problems 

1.  Solve  Problem  1  of  Art.  19  graphically  to  the  scale  of  1  inch  =  5  feet. 
Also  find  the  component  in  each  of  the  three  planes. 

2.  A  vector  20  feet  in  length  makes  an  angle  of  35  degrees  with  the  hori- 
zontal axis  east.     The  vector  is  in  a  plane  which  makes  an  angle  of  40  degrees 
with  the  horizontal  in  the  octant  above  the  horizontal  plane  and  north  of 
the  east  axis.     Find  the  components  of  this  vector  along  the  east  horizontal, 
the  north  horizontal,  and  the  vertical  axis. 

Figure  25  shows  the  graphical  method  of  finding  the  compo- 
nents of  a  vector  when  two  of  its  direction  angles  are  given.  The 


FIG.  24. 


FIG.  25. 


vector  of  length  I  makes  an  angle  a.  with  the  X  axis  and  an  angle 
|3  with  the  Y  axis.  In  the  V  plane  the  line  OP  of  length  /  is  first 
constructed  at  an  angle  a.  with  the  X  axis.  Its  projection  on  the 


CHAP.  II]        QUANTITY  AND  CALCULATIONS  25 

X  axis  is  the  component  Hx.  The  line  OP'  of  length  I  is  next 
drawn  at  an  angle  |8  with  the  Y  axis.  Its  projection  on  the  Y  axis 
is  the  component  V.  The  intersection  of  the  horizontal  line 
through  P'  with  the  vertical  line  through  P  gives  the  point  Q. 
This  point  is  the  projection  on  the  V  plane  of  the  terminus  of  the 
vector  in  space.  With  OQ  as  a  radius  and  0  as  a  center,  an  arc 
is  described  intersecting  the  X  axis  at  Q' '.  With  I  as  a  radius  and 
O  as  a  center,  a  second  arc  is  described  in  the  H  plane.  From  Q' 
a  line  is  drawn  in  the  H  plane  perpendicular  to  the  X  axis.  This 
line  intersects  the  arc  of  radius  I  at  the  point  P".  Since  OP" 
is  the  length  of  the  vector,  and  OQ'  is  the  length  of  its  component 
in  the  V  plane,  the  line  Q'  P",  which  completes  the  right  triangle 
OQ'P",  is  equal  to  the  component  Hz.  The  line  OR  of  length  H 
is  the  component  in  the  H  plane  and  the  line  OS  is  the  component 
in  the  profile  plane.  It  is  nob  necessary  to  find  these  last  two 
components,  but  it  is  sometimes  desirable  to  have  them. 

Problems 

3.  Solve  Problem  2  of.  Art.  20  graphically  to  the  scale  of  1  inch  =  10  feet. 

4.  Solve  Problem  3  of  Art.  20  graphically  to  the  scale  of  1  inch  =  10  feet. 

23.  Graphical  Determination  of  the  Vector  Sum. — To  find 
the  sum  of  several  vectors  in  space  by  graphical  methods,  each 
vector  is  first  resolved  into  components  along  the  three  axes  as 
in  Figs.  24  and  25.     To  avoid  confusion  it  is  often  advisable 
to  make  a  separate  drawing  for  the  resolution  of  each  vector. 
The  components  along  each  axis  are  added  graphically  and  laid 
off  along  the  corresponding  axis  of  the  final  diagram.    If  the  direc- 
tion of  the  vector  sum  is  desired  in  spherical  coordinates,  the 
construction  is  that  of  Fig.  24  in  the  inverse  order.     If  the  direc- 
tion angles  of  the  vector  sum  are  required,  they  may  be  obtained 
by  reversing  the  construction  of  Fig.  25. 

Problems 

1.  Solve  Problem  1  of  Art.  21  graphically  to  the  scale  of  1  inch  =  5  feet. 

2.  Solve  Problem  2  of  Art.  21  graphically  to  the  scale  of  1  inch  =  5  feet. 

3.  Solve  Problem  4  of  Art.  21  graphically  to  the  scale  of  1  inch  =  5  feet. 

24.  Product  of  Two  Vectors. — There  are  two  ways  of  multi- 
plying two  vectors.     The  result  of  one  method  is  a  vector  and  is 
called  the  vector  product  of  the  two  vectors.     The  result  of  the 


26  MECHANICS  [ART.  24 

other  method  is  a  mere  number  without  direction,  and  is  called 
scalar  product  of  the  two  vectors. 

If  the  length  of  vector  a  is  a  units  and  the  length  of  vector  b 
is  6  units,  and  if  the  angle  between  the  two  vectors  is  0, 

scalar  product  a.  b  =  ab  cos  6.  Formula  I. 

Since  b  cos  6  is  the  length  of  the  orthographic  projection  of  vec- 
tor b  upon  the  line  of  vector  a,  the  scalar  product  of  Wo  vectors 
may  be  defined  as  numerically  equal  to  the  product  of  the  magni- 
tude of  one  vector  multiplied  by  the  magnitude  of  the  projection 
of  the  other  vector  upon  its  direction.  Since 

ab  cos  0  =  ba  cos  0, 

either  vector  may  be  regarded  as  projected  upon  the  other. 
When  the  angle  between  the  two  vectors  is  zero,  their  scalar 
product  is  simply  the  product  of  their  magnitudes.  Scalar  mul- 
tiplication of  vectors,  when  they  are  in  the  same  direction,  is 
equivalent  to  multiplication  of  ordinary  arithmetic,  just  as 
addition  of  vectors  in  the  same  direction  is  equivalent  to 
addition  of  ordinary  arithmetic. 

The  work  done  by  a  force,  which  is  the  product  of  the  magni- 
tude of  the  force  multiplied  by  the  component  of  the  displace- 
ment in  the  direction  of  the  force,  is  an  example  of  a  scalar 
product. 

The  vector  product  of  two  vectors  a  and  b  is  defined  by  the 
equation 

vector  product  a  X  b  =  ab  sin  B.       Formula  II, 

The  vector  product  of  two  vectors  is  numerically  equal  to  the 
^  product  of  the  length  of   one  vector 

multiplied  by  the  component  of  the 
other  vector  perpendicular  to  its 
direction. 

The  vector  product  of  two  vectors 


/      i  A>     n  is  defined  as  a  vector  perpendicular 

^  to  the  plane  of  the  two  vectors  which 

are  multiplied  together.  In  the  right- 
handed  system  of  coordinates,  the  direction  of  the  angle  between 
the  vectors  and  the  direction  of  the  vector  product  bear  the 
same  relation  to  each  other  as  the  direction  of  rotation  and  motion 
of  a  right-handed  screw.  If  the  rotation  from  b  to  a  in  the  XY 
plane  of  Fig.  26,  I,  is  clockwise,  the  vector  product  bXa  is  along 


CHAP.  II]        QUANTITY  AND  CALCULATIONS  27 

the  Z  axis  away  from  the  observer.  If  the  rotation  from  a  to  b 
is  counter-clockwise,  the  vector  product  a  X  b  is  directed  along 
the  Z  axis  toward  the  observer. 

The  moment  of  a  force,  which  is  the  product  of  the  force 
multiplied  by  the  length  of  the  component  of  the  moment  arm 
perpendicular  to  the  direction  of  the  force,  is  an  example  of  a 
vector  product. 

There  are  several  ways  of  indicating  the  multiplication  of  two 
vectors.  The  most  convenient  one  is  that  used  by  Gibbs: 

Scalar  product  of  ab  =  a.b, 
which  is  read  a  dot  b  and  is  called  the  dot  product. 
Vector  product  of  ab  =  a  X  b, 

which  is  read  a  cross  b  and  is  called  the  cross  product. 

This  method  of  writing  the  vector  products  has  not  come  into 
general  use  and  is  not  employed  in  elementary  texts  on 
mechanics. 

Problems 

1.  Given  vector  a  =  6  feet  at  20°,  vector  b  =  4  feet  at  45°,  find  a.b, 
a  X  b,  and  b  X  a. 

Ans.    a.b  =  21.75;  a  X  b  =  10.14  ft.  toward  the  front  ;b  X  a  =  —  a  X  b. 

2.  A  vector  a  =  16.4  feet  at  0°,  a  vector  b  =  20.4  feet  at  25°,  and  a  vector 
c  =  18.2  feet  at  65°.     All  these  vectors  are  in  one  plane.     Find  a.b,  a.c, 
aXb,  aXc,  aXb  +  cXb,  aXb+aXc. 

Ans.    aXb+cXb  =  -  97.27  ft.;  aXb+aXc=  411.9  ft. 

25.  Summary. — The  magnitude  of  a  quantity  may  be  repre- 
sented by  a  number  or  by  the  length  of  a  line. 

A  quantity  which  has  magnitude  and  direction  is  called  a 
vector.  A  vector  is  represented  by  a  line. 

A  vector  in  a  plane  may  be  expressed  by  two  numbers.  One 
number  gives  its  length  and  the  other  gives  its  angle  with  a  known 
line.  A  vector  in  a  plane  may  also  be  expressed  by  means  of 
its  projections  on  two  lines. 

A  vector  in  space  may  be  expressed  by  three  numbers.  Its 
length  may  be  given  and  its  angle  with  two  lines  at  right  angles 
with  each  other.  These  angles  are  called  direction  angles  and 
their  cosines  are  direction  cosines  of  the  vector.  The  direction 
angle  with  the  third  axis  at  right  angles  to  the  plane  of  these  two 
lines  may  be  found  by  the  relation  of  analytic  geometry  that  the 
sum  of  the  squares  of  the  three  direction  cosines  is  equal  to  unity. 


28  MECHANICS  [Asa.  25 

The  direction  of  a  vector  in  space  may  also  be  found  by  means  of 
its  angle  with  one  axis  and  the  angle  which  the  plane  through  the 
vector  and  this  axis  makes  with  another  reference  plane  through 
the  axis.  This  method  is  equivalent  to  co-latitude  and  longitude 
and  is  sometimes  called  representation  by  spherical  coordinates. 
A  vector  in  space  may  also  be  expressed  by  means  of  its  projec- 
tions on  three  axis,  only  two  of  which  are  in  any  one  plane. 

The  sum  of  two  vectors  is  obtained  by  placing  the  origin  of  the 
second  vector  at  the  terminus  of  the  first  vector.  The  required 
sum  is  the  vector  which  joins  the  origin  of  the  first  vector  with 
the  terminus  of  the  second.  The  vector  sum  is  called  the  re- 
sultant vector. 

Vectors  which  added  together  form  a  given  vector  are  called 
components  of  the  given  vector.  The  most  important  compo- 
nents are  those  which  are  obtained  by  orthographic  projection. 

To  calculate  the  vector  sum  of  several  vectors  in  space,  each 
vector  is  first  resolved  into  its  components  along  the  three  co- 
ordinate axes  The  sum  of  the  components  along  any  axis  forms 
one  edge  of  a  rectangular  parallelepiped.  The  resultant  vector 
is  the  diagonal  of  this  parallelepiped.  If  the  vectors  are  in  a 
single  plane,  the  parallelepiped  is  reduced  to  a  rectangle,  and 
the  resultant  vector  is  the  hypotenuse  of  the  right-angled 
triangle  which  forms  one  half  of  this  rectangle. 

To  subtract  a  vector  add  an  equal  vector  in  the  opposite 
direction. 

There  are  two  kinds  of  products  of  vectors.  The  vector  product 
is  the  product  of  the  length  of  the  vectors  multiplied  by  the  sine  of 
the  angle  between  them.  This  product  is  a  vector  and  is  normal 
to  the  plane  of  the  given  vectors.  The  scalar  product  is  the 
product  of  the  length  of  the  vectors  multiplied  by  the  cosine  of 
the  angle  between  them.  This  p  oduct  is  a  scalar  quantity. 
It  has  magnitude  but  not  direction. 


CHAPTER  III 
APPLICATION  OF  FORCE 

26.  Tension,  Compression,  and  Shear. — Figure  27  shows  a  10- 
pound  mass  (a  so-called  10-pound  weight)  supported  by  a  cord, 
which  is  fastened  to  a  short  horizontal  beam  at  the  top.     The 
beam  pulls  upward  on  the  cord  at  the  top,  and  the  10- 
pound  mass  pulls  downward  at  the  bottom.     The  cord 

is  said  to  be  in  tension.  A  body  is  in  tension  when 
it  is  subjected  to  a  pair  of  equal  forces  which  are 
along  the  same  line,  opposite  in  direction,  and  away 
from  each  other.  A  body  in  tension  is  said  to  be 
under  tensile  stress. 

Figure"  28~  shows  a  50-pound  mass  resting  on  a  short  FlG<  27> 
block.  The  block  is  subjected  to  a  downward  push  of  50  pounds 
at  the  top,  and  an  upward  push  of  50  pounds  (in  addition  to 
its  own  weight)  from  the  support  at  the  bottom.  The  block  is 
in  compression  and  is  subjected  to  compressive  stress. 
A  body  is  in  compression  when  it  is  subjected  to  a 
pair  of  equal  forces  which  are  along  the  same  line, 
opposite  in  direction,  and  toward  each  other. 

In  Fig.  29,  the  weight  tends  to  slide  the  right 
FIG.  28.  portion  of  the  block  downward  relatively  to  the  left 
portion.  The  block  AB  is  said  to  be  in  shear  and  subjected  to 
shearing  stress.  A  body  is  in  shear  when  it  is  subjected  to  a 
pair  of  equal  forces  which  are  opposite  in  direction,  and  which 
act  along  parallel  planes. 

A  body  in  tension  is  lengthened  along  the  direction 
of  the  forces;  a  body  in  compression  is  shortened. 

27.  States   of   Matter. — Materials  exist  in  one  of 
three  forms,  solid,  liquid,  or  gaseous. 

A  block  of  steel  or  a  block  of  ice  is  an  example  of 
matter  in  the  solid  state.     A  solid  may  be  supported     FlG-  29- 
against  the  force  of  gravity  by  a  single  force  in  one  direction. 
A  block  of  ice  resting  on  a  table  is  supported  by  the  reaction  of 
the  table,  which  is  a  single  force  acting  upward.     A  flexible  cord, 
on  the  other  hand,  can  not  be  made  to  stand  on  the  lower  end, 

29 


30  MECHANICS  [ART.  28 

but  may  be  supported  on  the  side,  or  suspended  from  the  upper 
end,  and,  in  these  ways,  fulfill  the  condition  of  support  by  a  single 
force  in  one  direction. 

If  a  block  of  ice  is  heated  and  changed  into  the  liquid  state, 
it  can  no  longer  be  supported  by  a  single  force  upward,  but  must 
be  confined  on  all  sides  by  walls  which  exert  lateral  pressure. 

If  the  water  is  heated  still  further,  so  as  to  turn  it  into  steam, 
it  comes  into  the  gaseous  state.  It  must  now  be  confined  at 
the  top,  as  well  as  at  the  sides  and  bottom,  to  prevent  it  from 
expanding. 

A  solid  has  a  definite  form  and  a  definite  volume;  a  liquid  has  a 
definite  volume  but  not  a  definite  form;  a  gas  has  neither  a 
definite  form  nor  a  definite  volume. 

A  solid  can  resist  tension,  compression,  or  shear.  A  liquid 
offers  no  resistance  to  shear  and  will  support  practically  no  ten- 
sion. If  confined  so  as  to  prevent  lateral  shear,  a  liquid  or  gas 
will  resist  compression.  The  volume  of  a  gas  is  greatly  changed 
by  a  small  change  of  the  compression;  the  volume  of  a  liquid  is 
changed  very  little.  A  viscous  liquid  offers  some  resistance  to 
shear,  especially  when  the  force  is  applied  for  a  short  time.  Tar 
at  ordinary  temperatures,  and  steel  at  red  heat  are  viscous.  * 
Every  liquid  has  some  viscosity.  It  is,  however,  very  small  in 
the  case  of  water  or  alcohol,  and  relatively  large  in  heavy  oils. 
A  liquid  with  absolutely  no  viscosity  is  called  an  ideal  perfect 
fluid. 

28.  A  Rigid  Body.  —  A  solid  body,  which  suffers  little  change 
of  form  when  subjected  to  considerable  force,  is  called  a  rigid 
^    .....  ^  body.     All   bodies  are  elastic,   so  that 

0    '        —      there  is  some  change  in  form  or  dimen- 
sions  when  force  is  applied;  but  these 


..      ,. 

^     changes  are  frequently  so  small  as  to  be 
negligible,  except  for  measurements  re- 
,  quiring    the    greatest    accuracy.     Such 

bodies  are  regarded  as  rigid.     Fig.  30,  I, 

shows  a  beam  supported  at  the  middle  with  a  load  on  each  end. 
The  beam  is  somewhat  bent,  but  the  amount  of  bending  may  be 
so  small  that  the  distance  of  the  loads  from  the  vertical  line 
through  the  support  is  not  materially  changed.  Fig.  30,  II,  shows 
a  lighter  beam,  in  which  the  amount  of  bending  is  sufficiently 
great  to  change  materially  its  form  and  dimensions.  If,  how- 
ever, the  beam  be  considered  as  it  is  now  loaded,  it  may  be 


CHAP.  Ill] 


APPLICATION  OF  FORCE 


31 


regarded  as  a  rigid  body,  but  one  of  different  form  and  dimensions 
from  that  which  it  would  have  with  a  different  loading. 

In  elementary  calculations  of  Mechanics  no  allowance  is  made 
for  slight  elastic  deformations.  In  that  branch  of  Mechanics 
which  is  called  Strength  of  Materials  or  Mechanics  of  Materials, 
however,  these  deformations  are  taken  into  account. 

29.  A  Flexible  Cord. — Fig.  31,1,  shows  a  stiff  rope  supported  at 
the  middle.  This  rope  has  some  rigidity.  It  is  similar  to  a  very 
flexible  beam.  Fig.  31,  II,  shows  a  perfectly  flexible  cord.  The 


10  Ib. 
FIG.  32. 

cord  takes  the  form  of  the  support  at  the  top  and  hangs  vertically 
downward  at  the  ends.  An  ideal  flexible  cord  offers  no  resistance 
to  bending.  A  flexible*  cord  or  rope  can  exert  force  only  in  the 
form  of  tension  in  the  direction  of  its  length.  When  a  flexible 
cord  forms  a  part  of  a  structure  or  piece  of  apparatus  the  position 
and  direction  of  the  force  which  it  exerts  is  definitely  known. 

In  Fig.  32,  two  flexible  cords  are  attached  to  the  beam  AB. 
The  direction  of  the  tension  exerted  by  each  cord  on  the  beam  is 
shown  by  the  arrow;  the  position  of  the  force  in  each  is  known 
to  be  along  its  axis.  On  the  other  hand,  the  direction  of  the 
force  exerted  by  the  beam  at  B  can  not  be  determined  by  in- 
spection of  the  diagram,  but  must  be  calculated  from  the  direction 
and  magnitude  of  the  forces  in  the  cords. 

30.  Equilibrium. — Fig.  33  shows  a  10-pound 
mass  (an  ordinary  10-pound  weight)  supported 
by  a  flexible  cord.  If  the  mass  is  stationary 
with  respect  to  the  earth,  that  is,  if  it  is  not  swing- 
ing as  a  pendulum  or  vibrating  up  and  down  on 
an  elastic  support,  it  is  said  to  be  in  equilibrium. 
In  order  that  a  body  may  be  in  equilibrium,  the 
forces  which  act  on  it  from  other  bodies  must  balance.  Instead 
of  stating  that  the  body  is  in  equilibrium,  it  is  frequently  said 
that  the  forces  which  act  on  the  body  are  in  equilibrium.  A 


FIG.  33. 


32  MECHANICS  [ABT.  31 

body  in  equilibrium  is  not  necessarily  stationary.     It  may  be 
moving  with  constant  speed  in  a  straight  line. 

Two  forces  act  on  the  10-pound  mass  of  Fig.  33.  These  are 
the  pull  of  the  cord  directed  upward,  and  the  pull  of  gravity, 
acting  from  the  earth  through  the  ether,  directed  downward. 
These  two  forces  are  equal  in  magnitude;  that  is,  each  is  a  force 
of  10  pounds.  They  are  opposite  in  direction  and  are  exerted 
along  the  same  vertical  line. 

A  body  is  in  equilibrium  when  the  forces  in  any  direction  are 
balanced  by  the  forces  along  the  same  line  in  the  opposite 
direction. 

The  attraction  of  the  earth  and  the  pull  on  the  cord  in  Fig. 
33  do  not  constitute  action  and  reaction,  as  is 
often  erroneously  assumed.  There  are  two  sets 
of  action  and  reaction  involved  in  this  equi- 
librium. The  pull  of  the  earth  on  the  body  and 
the  pull  of  the  body  on  the  earth  form  one  action 
and  reaction.  The  pull  of  the  cord  on  the  body 
and  the  pull  of  the  body  on  the  cord  form  the 
other. 

In  Fig.  34,  the  knot  at  which  the  three  cords 
meet  may  be  regarded  as  the  bocjy  in  equilibrium.  The  down- 
ward pull  of  the  cord  from  the  10-pound  mass  is  balanced  by  the 
combined  upward  pulls  of  the  cords  from  the  spring  balances. 

31.  A  Smooth  Surface. — A  smooth  surface  is  one  which  offers 
no  resistance  to  the  motion  of  a  body  along  it.  The  only  force 
which  a  smooth  body  can  exert  is  normal  to  its  surface.  A  body 
in  motion  on  a  smooth,  horizontal  surface  will  continue  to  move 
for  an  indefinite  time  with  no  diminution  of  speed. 

No  surface  is  perfectly  smooth;  there  is  always  some  friction. 
Friction  is  a  force  at  the  surface  of  contact  of  two  bodies  which 
resists  the  motion  of  one  body  along  the  surface  of  the  other. 
The  friction  between  smooth  ice  and  a  polished  steel  runner,  or 
the  friction  between  a  metal  shaft  and  a  well  lubricated  bearing, 
is  small. 

Figure  35  represents  a  horizontal  surface  upon  which  a  body  B 
is  placed.  A  flexible  cord,  which  runs  over  a  smooth  pulley  and 
supports  a  weight  P  is  attached  to  B.  If  there  were  no  friction 
at  the  pulley,  the  tension  in  the  cord  at  B  would  be  equal  to  the 
weight  of  P.  With  some  friction,  the  tension  P'  is  slightly  less 
than  the  weight  of  P.  If  the  force  P'  parallel  to  the  surface  is 


CHAP.  Ill] 


APPLICATION  OF  FORCE 


33 


just  sufficient  to  start  the  body  in  motion,  this  force  is  said  to 
be  equal  to  the  starting  friction  between  the  body  and  the  surface 
upon  which  it  rests.  If  the  force  Pf  is  just  sufficient  to  keep  the 
body  moving  with  uniform  speed  after  it  has  been  started  by 
some  additional  force,  then  the  force  P'  is  said  to  be  equal  to  the 
moving  friction.  The  force  of 
friction  is  represented  in  Fig.  35 
by  the  single-barbed  arrow.  This 
arrow  indicates  that  the  fric- 
tion from  the  lower  surface  to 
the  body  B  is  directed  toward 
the  left.  This  is  opposite  to  the 
direction  of  the  force  P'.  The  FlG-  35- 

friction  of  the  body  B  upon  the  surface  below  it  is  directed  toward 
the  right. 

For  the  purpose  of  demonstrating  the  mechanics  of  ideal 
frictionless  surfaces,  the  moving  body  may  be  provided  with 
wheels  or  rollers,  as  in  Fig.  36.  This  figure  approximates 
closely  to  the  ideal  condition  of  a  smooth  bar  which  rests  on  a 
smooth  horizontal  floor  and  leans  against  a  smooth  vertical  wall. 
The  subject  of  friction  is  continued  in  Chapter  XII.  For  the 
present,  bodies  will  be  assumed  to  be  frictionless.  The  condition 
of  equilibrium  for  perfectly  smooth  bodies  is  approximately 
midway  between  the  limiting  conditions  of  equilibrium  for 
rough  bodies.  A  10-pound  mass  on  a  smooth  plane,  which  makes 
an  angle  of  30  degrees  with  the  horizontal,  may  be  held  in  equi- 
librium by  a  force  of  5  pounds  parallel  to  the  plane.  If  the  plane 
were  not  smooth,  the  body  would  still  be  held  in  equilibrium  by 
the  5-pound  force,  or  by  a  force  somewhat  greater  or  somewhat 
less  than  5  pounds.  If  the  coefficient  of  friction 
were  0.1,  as  in  Problem  10  of  Art.  112,  the  body 
would  be  held  by  any  force  parallel  to  the  plane 
between  the  limits  of  4.134  pounds  and  5.866  pounds. 
The  same  mass  may  be  held  on  a  smooth  plane  by 
a  horizontal  force  of  5.77  pounds;  and  may  be  held 
on  a  plane  for  which  the  coefficient  of  friction  is 
0.1  by  any  horizontal  force  between  the  limits  of  4.74  pounds 
and  7.19  pounds  (Problem  11  Art.  112).  In  the  first  case,  the 
force  which  holds  the  body  on  the  smooth  plane  is  exactly  mid- 
way between  the  limiting  forces  for  the  rough  plane;  in  the 
second  case,  it  deviates  a  little  from  the  median  value. 


FIG.  36. 


34  MECHANICS  (ART.  32 

32.  A  Smooth  Hinge. — Two  bodies  are  frequently  connected 
as  in  Fig.  37.  A  cylindrical  pin  passes  through  cylindrical  holes 
in  each  of  the  bodies  B  and  C.  The  pin  may  be  fixed  in  one  of  the 
bodies  but  must  be  free  to  turn  in  the  other.  This  form  of  connec- 
tion is  called  a  hinge.  One  body  can  rotate  relatively  to  the  other 
in  a  plane  which  is  perpendicular  to  the  axis  of  the  pin.  In  a  smooth 
hinge  or  pin-connection  the  force  between  the  pin  and  the  hollow 
cylinder  is  normal  to  the  curved  surfaces  at  the  line  of  contact. 
The  force  is,  therefore,  along  a  line  through  the  axes  of  the  pin 
and -the  hollow  cylinder,  and  in  a  plane  normal  to  these  axes. 

In  Fig.  37,  the  hollow  cylinder  in  body  C  is  drawn  much  larger 
than  the  pin  A.  In  practice  the  pin  is  made  to  fit  the  cylinder 
with  little  clearance. 


FIG.  37.  FIG.  38. 

A  body  with  a  smooth  hinge  at  each  end,  as  in  Fig.  38,  is 
called  a  link.  A  link  transmits  force  in  the  direction  of  the  line 
joining  the  two  hinges. 

33.  Resultant  and  Equilibrant. — In  Fig.  34,  the  upward  pull 
of  the  vertical  cord  is  replaced  at  the  knot  A  by  the  combined 
pull  of  the  cords  which  lead  to  the  spring  balances.  The  upward 
pull  in  the  vertical  cord  is  equal  to  the  resultant  of  the  forces  in 
the  other  two  cords.  The  downward  pull  of  the  vertical  cord 
at  the  knot,  which  balances  the  combined  upward  pull  of  the 
other  two  cords,  is  called  the  equilibrant.  The  resultant  of  a  com- 
bination of  forces  is  the  single  force  whose  effect  on  the  equilib- 
rium of  a  body  is  equivalent  to  that  of  the  combination  of  forces. 
The  equilibrant  of  a  combination  of  forces  is  the  single  force 
which  will  balance  these  forces  and  produce  equilibrium.  The 
resultant  of  a  set  of  forces  is  equal  and  opposite  to  their  equi- 
librant. Though  the  word  equilibrant  is  not  in  general  use,  the 
distinction  between  equilibrant  and  resultant  must  be  kept 
clearly  in  mind  to  avoid  confusion. 

In  Art.  29,  it  was  stated  that  the  force  in  each  flexible  cord  of 
Fig.  32  is  transmitted  along  the  axis  of  the  cord.  In  reality,  the 
force  is  transmitted  in  all  parts  of  the  cord.  The  resultant  of  all 


CHAP.  Ill]  APPLICATION  OF  FORCE  35 

these  forces  is  transmitted  along  the  geometrical  axis.  In  Art. 
30,  it  was  stated  that  the  pull  of  gravity  on  the  10-pound  mass  is 
vertically  downward.  In  reality,  every  particle  of  the  10  — 
pound  mass  is  attracted  by  every  particle  of  the  earth,  so  that 
some  of  the  force  is  nearly  horizontal.  The  resultant  of  all  these 
forces  is  vertically  downward. 

The  resultant  pull  of  gravity  on  a  body  passes  through  a  point 
which  is  called  the  center  of  gravity  or  center  of  mass  of  the  body. 

Any  system  of  forces  in  equilibrium  may  be  divided  into  two 
groups.  The  resultant  of  the  forces  of  one  group  forms  the 
equilibrant  for  the  resultant  of  the  forces  of  the  other  group. 
In  Fig.  34,  the  pull  from  the  left  spring  balance  may  be  regarded 
as  combined  with  the  downward  force  of  10  pounds.  The  resul- 
tant of  these  two  forces  is  equal  and  opposite  to  the  pull  from  the 
right  spring  balance.  In  like  manner,  the  pull  from  the  right 
spring  balance  may  be  combined  with  the  downward  force  of  10 
pounds.  The  resultant  of  these  two  forces  is  balanced  by  the 
pull  from  the  left  spring  balance.  The  force  in  any  one  of  the 
cords  of  Fig.  34  serves  as  the  equilibrant  which  balances  the 
resultant  of  the  forces  in  the  other  two  cords. 

The  forces  which  make  up  a  resultant  force  are  called  com- 
ponents of  the  resultant. 

In  the  calculation  of  problems  of  equilibrium,  it  is  often  possible 
to  use  the  resultant  of  a  number  of  forces,  instead  of  the  separate 
forces,  and,  in  this  way,  to  make  the  computations  more  simple, 
without  introducing  any  error  in  the  result. 

34.  The  Force  Circuit. — In  Fig.  27,  the  earth  pulls  downward  on 
the  10-pound  mass.     The  force  is  transmitted  from  the  earth 
through  the  ether.     The  10-pound  mass  pulls  down  on  the  hori- 
zontal beam.     The  force  is  transmitted  from  the  mass  to  the 
beam  in  the  form  of  tension  in  the  cord.     The  beam  pushes  down 
on  the  top  of  the  post  B  C.     The  force  is  transmitted  from  the 
top  of  the  cord  to  the  top  of  the  post  in  the  form  of  shear  in  the 
horizontal  beam.     The  force  is  transmitted  as  compression  from 
the  top  to  the  bottom  of  the  post.     It  is  next  transmitted,  toward 
the  left,  in  the  base  as  shear.     Finally  it  passes  to  the  earth  as 
compression,    and    the   resultant    compression   is    vertical    and 
directly   underneath    the    center   of   gravity   of  the   10-pound 
mass. 

35.  The  Free  Body. — In  a  problem  of  equilibrium  the  forces 
are  considered  which  act  on  some  definite  body  or  portion  of  a 


36  MECHANICS  [ART.  36 

body.  This  body  or  portion  of  a  body  is  called  the  free  body  in 
equilibrium.  In  Fig.  27,  the  10-pound  mass  may  be  taken  as  the 
free  body.  It  is  in  equilibrium  under  the  pull  of  the  earth 
downward  and  the  pull  of  the  cord  upward.  The  cord  may 
likewise  be  taken  as  the  free  body  in  equilibrium  under  the  pull 
of  the  10-pound  mass  and  the  pull  of  gravity  on  its  own  mass, 
both  of  which  are  balanced  by  the  upward  pull  of  the  beam  A  B. 
The  entire  system  of  Fig.  27  may  be  taken  as  a  free  body.  In 
this  case  the  weights  of  the  parts  of  the  system  form  one  set  of 
external  forces.  These  are  balanced  by  the  upward  reaction  at 
the  support. 

In  Fig.  35,  the  mass  B  may  be  regarded  as  the  free  body. 
The  forces  which  act  on  it  are,  (1)  the  force  of  gravity  downward, 
(2)  the  vertical  reaction  of  the  plane  upward,  (3)  the  horizontal 
friction  of  the  supporting  plane  toward  the  left,  and  (4)  the 
horizontal  pull  of  the  rope  toward  the  right.  The  force  of  gravity, 
or  weight,  is  represented  by  the  arrow  marked  W.  The  upward 
reaction  of  the  plane  is  represented  by  the  arrow  marked  N. 
Generally  the  arrow  which  represents  the  weight  is  drawn  down- 
ward from  the  body;  in  this  case  it  is  placed  above  the  body  and 
drawn  downward  to  it.  This  is  done  in  order  to  avoid  confusion 
with  the  upward  reaction  N,  which  would  fall  on  the  same  line 
if  both  were  drawn  below  the  body.  The  friction  from  the 
supporting  plane  to  the  body  is  represented  by  the  single-barbed 
arrow  F;  and  the  pull  from  the  cord  by  the  arrow  P'.  The 
arrows  must  show  the  direction  of  the  force  exerted  by  the  other 
bodies  or  parts  of  a  body  upon  the  free  body.  If  B  is  the  free 
body,  F  is  toward  the  left  and  N  is  upward.  If  the  supporting 
plane  were  under  consideration  as  the  free  body,  then  the  friction 
F  from  B  to  the  plane  would  be  toward  the  right,  and  the  normal 
N  from  B  to  the  plane  would  be  downward. 

In  considering  a  problem  of  equilibrium,  it  is  absolutely 
necessary  to  decide  what  body  or  portion  of  a  body  is  to  be 
taken  as  the  free  body,  and  then  to  designate  all  the  forces  which 
act  on  that  free  body. 

36.  Application  of  Forces.— In  Fig.  34,  the  three  cords  meet 
at  a  point  at  the  knot  A.  Forces  which  meet  at  a  point  are 
concurrent.  All  these  cords  lie  in  one  vertical  plane.  Forces 
in  the  same  plane  are  coplanar.  Fig.  34  is  an  illustration  of 
concurrent,  coplanar  forces. 


CHAP.  Ill] 


APPLICATION  OF  FORCE 


37 


In  Fig.  39,  the  three  forces  do  not  meet  at  a  point.  They  are 
non-concurrent.  All  these  forces  lie  in  one  plane.  Fig.  39  is  an 
illustration  of  non-concurrent,  coplanar  forces. 


FIG.  39. 


FIG.  40. 


In  Fig.  40,  there  are  four  forces  which  meet  at  the  point  B. 
They  are  concurrent.  They  do  not  all  lie  in  one  plane.  They 
are  non-coplanar.  Fig.  40  is  an  illustration  of  concurrent,  non- 
coplanar  forces. 


In  Fig.  41,  the  forces  which  act  on  the  horizontal  bar  B  do  not 
meet  at  one  point  and  do  not  lie  in  one  plane.  Fig.  41  is  an 
illustration  of  non-concurrent,  non-coplanar  forces. 

These  four  classes  of  the  application  of  forces  will  be  con- 
sidered separately  in  the  chapters  which  follow. 

37.  Resultant  of  Concurrent  Forces. — A  force  has  magnitude 
and  direction  and  a  definite  position.  A  vector  has  magnitude 
and  direction,  so  that  it  is  natural  to  assume  that  a  force  may  be 
represented  by  a  vector  and  that  the  resultant  of  two  concurrent 
forces  may  be  represented  by  their  vector  sum.  Experiments 
show  that  this  assumption  is  true,  so  that  it  may  be  stated  as  an 
axiom,  which  has  been  amply  verified  by  measurements:  // 


38  MECHANICS  [ART.  38 

two  concurrent  forces  are  represented  by  vectors,  the  resultant  of  the 
forces  is  represented,  in  direction  and  magnitude,  by  the  vector  sum 
of  these  two  vectors.  The  two  vectors  and  their  resultant  form  a 
triangle,  Fig.  42,  II,  which  is  called  the  force  triangle.  The 
resultant  force  passes  through  the  point  at  which  the  two  forces 
are  concurrent, — the  point  0  of  Fig.  42,  I. 

Since  the  resultant  of  two  concurrent 
forces  may  be  represented  by  their 
vector  sum,  the  resultant  of  three  con- 

^,         -r        -.  j^  r        current  forces  may  be  represented  by 
\^  u  the    vector    sum    obtained    by    com- 

FIG  42  bining    the   resultant   of  the  first  two 

forces  with  the  third  force.     The  third 

force  may  not  be  in  the  plane  of  the  first  two  forces.  When  the 
three  forces  are  all  in  one  plane,  it  is  not  necessary  to  draw  the 
resultant  of  the  first  two  forces  and  then  to  draw  the  third 
force.  All  three  forces  may  be  added  at  one  time,  as  in  Fig.  43. 
The  figure  thus  obtained  is  called  the  force  polygon. 

The  vector  sum  of  two  vectors  a  and  b  may  be  drawn  by  adding 
vector  b  to  vector  a,  or  by  adding  vector  a  to  vector  b,  as  shown 
in  Fig.  11.  If  the  addition  is  made  in  both  orders,  each  starting 
from  the  same  origin,  the  figure  obtained  is  a  parallelogram.  For 
this  reason,  the  method  of  finding  the  resultant  of  two  forces  by 
means  of  their  vector  sum  is  often  called  the  parallelogram  law. 
It  is,  however,  entirely  unnecessary  to  draw  the  parallelogram. 
The*  single  triangle,  which  is  one-half  of  the  parallelogram,  is 
sufficient.  Moreover,  the  use  of  par- 
allelograms causes  confusion  when 
three  or  more  forces  are  involved  in 
the  problem.  It  is  best,  therefore,  to 
consider  the  force  triangle  and  the 
force  polygon,  and  to  forget  the 
parallelogram. 

The  process  of  finding  the  resultant  of  two  or  more  forces  is 
called  the  composition  of  forces,  and  the  forces  which  make  up 
the  resultant  are  called  its  components. 

38.  Resolution  of  Forces. — The  process  of  finding  the  compo- 
nents of  a  force  is  resolution  of  the  force.  The  force  is  said  to  be 
resolved  into  its  components.  The  most  important  case  of  resolu- 
tion is  that  in  which  the  components  are  at  right  angles  to  each 
other.  There  may  be  two  components  at  right  angles  to  each 


CHAP.  Ill]  APPLICATION  OF  FORCE  39 

other  in  the  same  plane,  or  three  components,  any  one  of  which 
is  perpendicular  to  the  plane  of  the  other  two.  When  the  word 
component  is  used,  unless  otherwise  stated,  a  component  of  this 
kind  is  meant.  Frequently  only  one 
component  is  considered.  Fig.  44 
shows  a  force  P  at  an  angle  with  the 
line  A B.  The  component  of  the 
force  P  in  the  direction  of  the  line 
AB  is  equal  in  length  to  the  ortho- 
graphic projection  upon  AB  of  the  vector  which  represents 
the  force. 

The  process  of  finding  the  orthographic  component  of  a  force 
in  a  given  direction  is  called  resolution  in  that  direction. 

39.  Work. — Figure  45  represents  a  body  B  acted  on  by  a  force  P. 
The  force  is  applied  at  the  point  C,  which  is  called  its  point  of 
application.  The  force  may  be  exerted  through  a  flexible  cord, 
which  is  tied  to  the  body  at  the  point  of  application.  The  body 
moves  to  a  second  position  while  the  force  continues  to  act  in  the 

same  direction.  The  point 
C  moves  a  distance  of  s 
units  of  length.  This  is 
the  displacement  of  the 
point  of  application.  If  the 
body  does  not  rotate,  the 
displacement  of  all  parts  is  the  same.  If,  however,  there  is  any 
rotation,  the  displacement  of  the  point  of  application  may  be 
different  from  that  of  other  parts  of  the  body. 

If  the  displacement  of  the  point  of  application  of  the  force 
makes  an  angle  a  with  the  direction  of  the  force,  the  work  done 
by  the  force  is  defined  by  the  equation', 

work  =  P  cos  a  s  Formula  III. 

Since  P  cos  a  is  the  component  of  the  force  in  the  direction 
of  the  displacement,  the  definition  of  work  may  be  stated  as 
follows :  The  work  of  a  force  is  the  product  of  the  displacement  of 
its  point  of  application  multiplied  by  the  component  of  the  force  in 
the  direction  of  the  displacement. 

P  cos  a  s  =  P  s  cos  a. 

From  the  second  member  of  this  equation,  work  may  be  defined 
as  the  product  of  the  force  multiplied  by  the  component  of  the 
displacement  in  the  direction  of  the  force. 


40 


MECHANICS 


[ART.  40 


, 


If  the  force  is  expressed  in  pounds  and  the  displacement  in 
feet,  the  work  is  in  foot-pounds.  Other  units  of  work  are  inch- 
pounds,  gram-centimeters,  kilogram-meters,  and  ergs. 

Both  force  and  displacement  are  vectors.  The  product  of  two 
vectors  may  be  a  third  vector,  or  may  be  a  scalar  quantity, 
which  has  magnitude  but  not  direction.  Work  is  a  scalar  prod- 
uct, and  is  not  a  vector. 

40.  Classes  of  Equilibrium. — There  are  three  kinds  of  equilib- 
rium. These  are  called  stable,  unstable,  and  neutral  or  indifferent. 

Figure  46  shows  three  cases  of  stable  equilibrium.  When  a  body 
in  stable  equilibrium  is  displaced  slightly  from  its  position  of 


III 


equilibrium  by  an  additional  force,  it  will  return  to  that  position 
of  equilibrium  when  the  additional  force  ceases  to  act.  In  Fig. 
46,  I,  the  body  is  hung  on  a  flexible  cord.  The  body  is  in  stable 
equilibrium  in  position  A.  The  cord  is  vertical  and  the  center 
of  mass  is  directly  under  the  point  of  support.  If  an  additional 
force  is  applied,  which  deflects  the  body  to  position  B,  it  will 
return  to  position  A  when  the  additional  force  is  removed.  In 
the  position  of  equilibrium  the  forces  which  act  on  the  body  are 
its  weight  and  the  tension  in  the  cord.  When  the  body  is 
moved  from  position  A  to  position  B,  no  work  is  done  by  the 
tension  in  the  cord,  since  the  displacement  is  perpendicular  to 
the  direction  of  the  force ;  negative  work  is  done  by  the  weight, 
since  the  upward  component  of  the  displacement  is  opposite  the 
direction  of  the  force  of  gravity.  When  a  body  in  stable  equi- 


CHAP.  Ill] 


APPLICATION  OF  FORCE 


41 


librium  is  displaced  slightly,  the  original  forces  which  produce 
equilibrium  do  negative  work,  and  positive  work  must  be  done  by 
the  additional  forces  to  cause  the  displacement. 

Figure  46,  II,  shows  a  body  on  a  smooth,  curved  surface  which  is 
concave  upward.  The  forces  which  produce  equilibrium  are  the 
weight  of  the  body  and  the  normal  reaction  of  the  surface.  If 
the  surface  is  that  of  a  sphere  or  circular  cylinder,  the  conditions 
are  the  same  as  those  of  a  body  suspended  by  a  flexible  cord. 

Figure  46,  III,  shows  a  body  suspended  from  a  smooth  hinge 
and  attached  to  a  cord  which  runs  over  a  smooth  pulley  and 
carries  a  second  body  Q.  If  the  body  is  displaced  toward  the 
right,  the  weight  W  does  negative  work  and  the  pull  in  the  cord 
does  positive  work.  It  may  be  shown  that  the  negative  work  is 
the  greater,  and  that  the  equilibrium  is  stable. 

Figure  47  shows  three  cases  of  unstable  equilibrium.  Figure 
47,  I,  shows  a  body  on  a  smooth  curved  surface  which  is  concave 


W 


FIG.  47. 

downward.  If  the  body  is  moved  slightly  in  either  direction 
from  the  position  of  equilibrium,  it  will  not  return,  but  will 
continue  to  move  in  that  direction.  The  vertical  component  of 
the  displacement  is  in  the  same  direction  as  the  force  of  gravity; 
hence  its  weight  does  positive  work  when  it  is  moved  from  the 
position  of  equilibrium. 

Figure  47,  II,  shows  a  body  supported  on  a  smooth  hinge  which 
directly  under  its  center  of  mass.     The  motion  is  the  same  as 


42 


MECHANICS 


[AKT.  40 


that  of  Fig.  47,  I,  when  the  surface  is  cylindrical.  The  body  of 
Fig.  47,  II,  will  rotate  about  the  hinge  until  its  center  of  mass  is 
directly  under  the  point  of  support.  It  will  then  be  in  stable 
equilibrium  in  the  position  of  Fig.  46,  I.  In  like  manner,  the 
body  of  Fig.  47,  I,  will  move  to  the  position  of  Fig.  46,  II,  pro- 
vided the  surface  is  continuous,  and  there  is  some  arrangement  to 
keep  it  from  falling  off  when  below  the  center. 

Figure  47,  III,  is  similar  to  Fig.  46,  III,  but  the  equilibrium  is 
unstable.  If  displaced,  it  will  rotate  to  the  position  of  stable 
equilibrium  of  Fig.  46,  III. 


FIG.  48. 

Figure  48  shows  two  cases  of  neutral  equilibrium.  The  body  re- 
mains in  equilibrium  in  any  position.  Figure  48,  I,  represents  a 
body  on  a  smooth,  horizontal  plane  surface.  No  work  is  done  when 
it  is  displaced,  as  the  weight  and  the  normal  reaction  of  the  sur- 
face are  both  perpendicular  to  the  direction  of  the  displacement. 

Figure  48,  II,  shows  a  body  on  a  smooth  inclined  plane.  The 
body  is  supported  by  a  cord  which  runs  parallel  to  the  plane  and 
exerts  a  constant  pull  Q.  If  the  body  is  displaced  on  the  plane, 
the  positive  work  of  the  force  Q  is, equal  to  the  negative  work  of 
the  weight.  If  the  body  is  displaced  down  the  plane,  the  positive 
work  of  the  weight  is  equal  to  the  negative  work  of  the  force  Q. 


A^r 


CHAPTER  IV 


CONCURRENT  CO-PLANAR  FORCES 

41.  Resultant. — If  the  direction  and  magnitude  of  each  of 
several  forces  is  represented  by  a  vector,  the  direction  and  mag- 
nitude of  the  resultant  of  these  forces  will  be  represented  by  the 
vector  sum  of  these  vectors.  As  stated  in  Art.  37,  this  statement 
may  be  regarded  as  an  axiom  which  has  been  amply  verified  by 
experiment.  If  these  forces  are  concurrent  at  a  given  point, 
their  resultant  will  pass  through  this  point. 

In  Fig.  49,  I,  two  flexible  cords  in  a  vertical  plane  are  attached 
to  a  beam  at  a  point  A.  Each  cord  runs  over  a  smooth  pulley. 


II 


III 


FIG.  49. 


The  left  cord  makes  an  angle  of  35  degrees  to  the  left  of  the 
vertical,  and  supports  a  mass  of  8  pounds;  the  right  cord  makes 
an  angle  of  50  degrees  to  the  right  of  the  vertical,  and  supports 
a  mass  of  12  pounds.  If  the  pulleys  are  frictionless,  the  tension 
of  each  cord  at  A  will  be  equal  to  the  mass  which  it  supports. 
Fig.  49,  II,  is  the  space  diagram  representing  the  direction  and 
location  of  these  forces,  which  are  concurrent  at  A .  Figure  49,  III, 
is  the  force  polygon,  which,  in  this  case,  is  a  triangle.  To  con- 
struct the  force  triangle,  point  0  is  selected  as  the  origin  and  a 
line  is  drawn  through  this  point  parallel  to  the  direction  of  the 
force  of  12  pounds  in  the  space  diagram.  A  length  of  12  units, 
from  0  to  C,  is  measured  in  this  line.  The  point  C  is  the  terminus 
of  the  vector  which  represents  the  12-pound  force.  From  C 
as  the  origin,  a  second  vector,  8  units  in  length,  is  drawn  parallel 
to  the  direction  of  the  force  of  8  pounds  in  the  space  diagram. 

43 


44  MECHANICS  [ART.  42 

The  direction  and  magnitude  of  the  resultant  of  these  two  forces 
is  given  by  the  line  0  D  which  is  drawn  from  the  origin  of  the  first 
vector  to  the  terminus  of  the  second  vector.  Finally,  through  the 
point  of  application  of  the  forces  on  the  space  diagram,  a  broken 
line  is  drawn  parallel  to  the  direction  of  the  resultant  OD  of  the 
force  triangle.  This  broken  line  gives  the  location  of  the  line  of 
action  of  the  resultant  force. 

Problems 

1.  Find  the  magnitude  and  direction  of  the  resultant  of  the  following 
forces  in  a  horizontal  plane:  20  pounds  north  75  degrees  east,  15  pounds 
north  20  degrees  east,  and  8  poun,ds  north  25  degrees  west.     Solve  graphi- 
cally to  the  scale  of  1  inch  =  5  pounds. 

2.  Two  ropes  in  a  vertical  plane  are  attached  to  a  fixed  point.     One  rope, 
which  supports  a  mass  of  40  pounds  on  the  free  end,  runs  over  a  smooth 
pulley  located  16  feet  higher  than  the  fixed  point,  and  7  feet  to  the  left  of 
the  vertical  line  through  it.   »The  second  rope,  which  supports  a  mass  of 
35  pounds,  runs  over  a  smooth  pulley  located  8  feet  higher  than  the  fixed 
point,  and  15  feet  to  the  right  of  the  vertical  line  through  it.     Construct 
the  space  diagram  to  the  scale  of  1  inch  =  5  feet.     Then  construct  the  force 
triangle  to  the  scale  of  1  inch  =  10  pounds.     Measure  the  resultant  in  the 
force  triangle  and  express  its  magnitude  in  pounds  and  its  angle  with  the 
vertical  in  degrees  and  minutes.     Draw  a  broken  line  in  the  space  diagram 
to  show  the  line  of  action  of  the  resultant  force. 

42.  Calculation  of  Resultants  and  Components. — The  re- 
sultant of  two  forces  may  be  calculated  by  means  of  the  solution 
of  the  force  triangle. 

Problems 

1.  Find  the  resultant  of  a  force  of  24  pounds  and  a  force  of  30  pounds 
which  makes  an  angle  of  35  degrees  with  the  direction  of  the  24-pound  force. 
Construct  the  force  triangle  and  solve  for  the  magnitude  of  the  resultant 
by  means  of  the  law  of  cosines.     After  the  magnitude  is  found,  calculate 
the  angle  by  means  of  the  law  of  sines.     Check  all  results  by  means  of  the 
projections  of  the  forces  upon  the  line  of  action  of  the  resultant. 

2.  A  force  of  234  pounds  is  north  24  degrees  west,  and  a  concurrent  force 
of  256  pounds  is  north  17  degrees  east.     Construct  the  force  triangle  and 
solve  for  the  direction  of  the  resultant  by  means  of  the  ratios  of  the  sum 
and  difference  of  the  sides  and  the  tangents  of  one-half  the  sum  and  one- 
half  the  difference  of  the  angles  opposite  these  sides.     After  all  the  angles 
are  found,  calculate  the  side  which  represents  the  resultant  by  means  of 
the  law  of  sines. 

3.  A  force  P  makes  an  angle  of  43  degrees  to  the  right  of  the  vertical. 
A  force  Q,  concurrent  with  it,  has  a  magnitude  of  84.26  pounds.     The 
resultant  of  these  forces  is  a  vertical  force  of  116.45  pounds.     Find  the 


CHAP.  IV]     CONCURRENT  CO-PLANAR  FORCES  45 

direction  of  the  force  Q  and  the  magnitude  of  the  force  P,  using  logarithms. 
Construct  the  force  triangle  to  the  scale  of  1  inch  =  20  pounds  and  compare 
with  the  calculated  results. 

4.  The  resultant  of  two  forces  of  16  pounds  and  24  pounds  in  a  horizontal 
plane  is  a  force  of  34  pounds  north  10  degrees  east.     Find  the  direction  of 
the  forces  by  means  of  the  formula  for  the  tangent  of  the  half  angle. 

When  it  is  desired  to  find  the  resultant  of  more  than  two  forces, 
the  method  of  calculation  by  means  of  triangles  is  laborious. 
To  solve  such  problems  each  force  is  resolved  along  the  direction 
of  two  axes  which  are  at  right  angles  to  each  other.  The  sum 
of  the  components  of  all  these  forces  along  one  axis  is  the  compo- 
nent of  the  resultant  along  that  axis,  and  the  sum  of  the  compo- 
nents of  all  these  forces  along  the  other  axis  is  the  component  of  the 
resultant  along  that  axis.  These  two  components  of  the  resul- 
tant form  two  sides  of  a  right-angled  triangle  of  which  the  resul- 
tant is  the  hypotenuse.  When  the  resultant  of  several  forces  is 
to  be  found,  the  work  may  be  arranged  in  tabular  form,  as  was 
shown  in  Problem  4  of  Art.  17. 

Problems 

5.  A  horizontal  force  of  25  pounds  is  north  37  degrees  east,     Find  the 
east  and  north  components. 

Ans.     East  component  =  15.05  Ib. ;  north  component  =  19.97  Ib. 

6.  A  horizontal  force  of  46  pounds  is  north  22  degrees  west.     Find  the 
component  north  and  the  component  east. 

Ans.    East  component  =  —17.23  Ib.;  north  component   =  42.65  Ib. 

7.  A  force  of  24  pounds  makes  an  angle  of  20  degrees  with  the  horizontal. 
A  concurrent  force  of  30  pounds  in  the  same  vertical  plane  makes  an  angle 
of  64  degrees  with  the  horizontal  on  the  same  side  of  the  vertical.     Find  the 
sum  of  the  horizontal  and  the  sum  of  the  vertical  components. 

Ans.     H  =  35.70  Ib.;  V  =  35.17  Ib. 

8.  Find  the  direction  and  magnitude  of  the  resultant  of  the  two  forces  of 
Problem  7. 

9.  Solve  Problem  8  by  resolving  along  the  line  of  the  force  of  24  pounds 
and  along  a  line  perpendicular  to  the  force  of  24  pounds  in  the  same  vertical 
plane. 

10.  Find  the  direction  and  magnitude  of  the  resultant  of  14  pounds  north 
27  degrees  east,  15  pounds  north  35  degrees  west,  and  18  pounds   south 
56  degrees  west.     Resolve  east  and  north  then  check  by  resolutions  along 
some  other  pair  of  axes. 

11.  Solve  Problem  1  of  Art.  41  by  means  of  resolutions,  and  compare 
the  results  with  the  graphical  solution. 

12.  Solve  Problem  2  of  Art.  41  and  compare  the  results  with  the  graphical 
solution. 


46  MECHANICS  [ART.  43 

13.  Find  the  direction  and  magnitude  of  the  resultant  of  24.2  pounds  at 
20  degrees,  17.8  pounds  at  65  degrees,  22.5  pounds  at  110  degrees,  12.6 
pounds  at  165  degrees,  and  31.4  pounds  at  234  degrees.  Resolve  along 
0  degrees  and  90  degrees  and  check  by  resolutions  along  20  degrees  and  110 
degrees. 

43.  Equilibrium. — When  a  body  is  in  equilibrium  under  the 
action  of  two  forces,  these  forces  are  equal,  opposite,  and  along 
the  same  line.  If  a  body  is  in  equilibrium  under  the  action  of 
three  forces,  the  resultant  of  any  two  of  these  forces  must  be 
equal  and  opposite  the  third  force,  and  must  act  along  the  same 

line.     Figure    50  shows  three 
flexible    cords   attached    to  a 
point.     The   resultant   of  the 
forces  P  and  Q  is  the  force  R. 
The  direction  of  R  is  opposite 
II        ^     ill        the  direction  of  the  third  force 
5Q  T.     Figure  50,  II,   shows  the 

force  triangle  for  the  resultant. 

Figure  50,  III,  shows  the  triangle  for  the  three  forces  in  equi- 
librium. The  only  difference  between  Fig.  50,  II,  and  Fig.  50, 
111,  is  the  direction  of  one  line.  The  resultant  in  Fig.  50,  II, 
extends  from  the  origin  of  the  vector  P  to  the  terminus  of  the 
vector  Q.  The  equilibrant  in  Fig.  50,  III,  extends  from  the 
terminus  of  sector  Q  to  the  origin  of  vector  P.  This  last  vector 
T  is  called  the  closing  line. 

When  a  body  is  in  equilibrium  under  the  action  of  several 
concurrent  forces,  the  resultant  of  all  the  forces  is  zero,  as  may 
be  seen  from  Fig.  50,  III.  The  arrows  which  represent  the 
direction  of  the  forces  follow  each  other  around  the  force  diagram 
from  the  origin  of  the  first  vector  to  the  terminus  of  the  last 
vector.  The  arrow  on  one  side  of  each  angle  of  the  diagram 
points  toward  the  angle  and  the  arrow  on  the  other  side  of  the 
angle  points  away  from  it.  The  terminus  of  the  last  vector 
coincides  with  the  origin  of  the  first.  It  is  customary  to  say 
that  the  force  polygon  closes  when  equilibrium  exists. 

To  solve  a  problem  of  equilibrium  of  concurrent  forces, 
that  part  of  the  force  polygon  which  represents  the  known 
forces  is  constructed  first.  If  there  is  only  one  unknown 
force,  it  is  represented  in  the  force  polygon  by  the  closing  line, 
which  is  drawn  from  the  terminus  of  the  last  known  force  to  the 
origin  of  the  first  one.  If  there  are  two  forces  of  known  direction 


CHAP.  IV]     CONCURRENT  CO-PLANAR  FORCES 


47 


but  of  unknown  magnitude,  a  line  is  drawn  through  the  terminus  of 
the  last  known  vector  in  the  direction  of  one  of  these  unknown 
forces,  and  a  line  is  drawn  through  the  origin  of  the  first  known 
vector  in  the  direction  of  the  other  unknown  force.  The  length 
of  each  of  these  lines  to  their  point  or  intersection  represents  the 
required  force. 

Example 

A  40-pound  mass  rests  on  a  smooth  inclined  plane,  which  makes  an  angle 
of  35  degrees  with  the  horizontal.  It  is  supported  by  a  horizontal  push  of 
12  pounds,  and  a  pull  P  parallel  to  the  plane.  Find  the  force  P  and  the 
normal  reaction. 

Figure  51,  I,  is  the  space  diagram  showing  the  direction  of  all  the  forces. 
The  mass  of  40  pounds  is  the  free  body  in  equilibrium.     The  arrows  show 
the    direction    of  the   forces 
which  act  from  other  bodies 
on  the  free  body.     The  reac- 
tion of  the  plane  is  upward  at 
an  angle  of  35  degrees  with 
the  vertical.     The  pull  of  the 
earth  on  the  free  body  is  a 
downward  force  of  40  pounds. 
The  length  of  the  lines  in  this 
diagram  is  not  necessarily  pro- 
portional to  the  magnitude  of  FlG    51 
the  forces. 

Figure  51,  II,  is  the  force  polygon.  The  known  forces  are  40  pounds  down- 
ward and  12  pounds  horizontal  toward  the  right.  As  the  first  step  in  the 
construction  of  the  force  polygon  lay  off  AB  40  units  in  length.  From  B 
draw  the  horizontal  line  BC  12  units  in  length.  The  arrow  in  AB  is  down- 
ward toward  B,  and  the  arrow  in  BC  is  to  the  right,  directed  away  from  B. 
Through  C  draw  a  line  parallel  to  the  force  P  of  the  space  diagram,  and 
through  A  draw  a  line  parallel  to  the  direction  of  the  normal  reaction.  Ex- 
tend these  two  lines  till  they  intersect  at  D.  The  length  of  CD  measures 
the  tension  parallel  to  the  inclined  plane,  and  the  length  of  DA  represents 
the  normal  reaction  of  the  plane  against  the  40-pound  mass.  The  arrow 
in  BC  is  toward  C.  The  arrow  in  CD  must  by  away  from  C  toward  D;  the 
corresponding  direction  in  the  space  diagram  is  up  the  plane.  The  arrow 
in  DA  is  from  D  toward  A. 

Problems 

1.  Solve  the  example  above  graphically  to  the  scale  of  1  inch  =  10  pounds, 
with  the  horizontal  force  20  pounds  instead  of  12  pounds. 

2.  A  20-pound  mass  on  a  35  degree  inclined  plane  is  held  in  equilibrium 
by  a  horizontal  push  of  5  pounds,  and  by  a  pull  at  an  angle  of  40  degrees 
with  the  horizontal  and  an  angle  of  5  degrees  with  the  plane.     Solve  for 
this  pull  and  the  normal  reaction  to  the  scale  of  1  inch  =  5  pounds. 


i  sib. 


48 


MECHANICS 


[AET.  44 


3.  A  50-pound  mass  is  supported  by  3  cords  in  the  same  vertical  plane. 
One  cord  makes  an  angle  of  45  degrees  to  the  left  of  the  vertical.     The 
second  cord  makes  an  angle  of  30  degrees  to  the  right  of  the  vertical.     The 
third  cord  runs  horizontally  toward  the  right  and  exerts  a  pull  of  10  pounds. 
Find  the  tension  in  the  first  two  cords,  graphically,  to  the  scale  of  1  inch  = 
10  pounds. 

4.  A  40-pound  mass  is  supported  by  two  cords,  one  of  which  makes  an 
angle  of  35  degrees  to  the  right  of  the  vertical,  and  the  other  exerts  a  pull  of 
25  pounds.     Find  the  direction  of  the  second  cord  and  the  tension  in  the 
first  one,  graphically,  to  the  scale  of  1  inch  =  10  pounds.     There  are  two 
solutions.     Are  both  solutions  possible  with  cords? 

6.  A  40-pound  mass  is  supported  by  a  cord  and  a  rod  hinged  at  the  ends. 
One  of  these  makes  an  angle  of  24  degrees  to  the  right  of  the  vertical,  and 
the  other  makes  an  angle  of  85  degrees  to  the  right  of  the  vertical.  Find 
the  tension  in  the  cord  and  the  compression  in  the  rod. 

I* 

44.  Equilibrium  by  Resolutions. — The  force  diagram  for  con- 
current, coplanar  forces  in  equilibrium  is  a  closed  polygon. 
If  a  closed  polygon  is  projected  upon  any  line  in  its  plane,  the 
sum  of  the  positive  projections  is  equal  to  the  sum  of  the  nega- 
tive projections,  and  the  total  projection  is  zero.  It  follows, 
therefore,  that  when  a  set  of  forces  are  in  equilibrium,  the  sum  of 


FIG.  52. 

the  components  of  these  forces  along  any  direction  is  zero. 
Fig.  52,  I,  is  the  space  diagram  for  Problem  3  of  Art.  43  and 
OA  SCO,  in  Fig.  52,  II,  is  the  force  polygon.  Figure  52,  III,  shows 
the  projections  of  the  sides  of  the  force  pfcftygon  upon  a  line  at  an 
angle  of  10  degrees  with  the  horizontal.  The  projections  O'A', 
A'B',  and  B'C',  extend  from  left  to  right.  The  projection  C'O', 
extends  from  right  to  left  to  the  point  of  beginning. 

To  solve  a  problem  of  equilibrium  of  concurrent  forces  by  resolu- 
tions, it  is  not  necessary  to  draw  the  force  polygon,  as  the  com- 
ponents may  be  calculated  from  the  space  diagram.  However, 


CHAP.  IV]     CONCURRENT  CO-PLANAR  FORCES 


49 


always  draw  the  space  diagram  and  mark  all  the  forces  which  act 
on  the  free  body. 

Figure  53  shows  the  space  diagram  for  four  forces  PI,  P2,  PS, 
and  P4,  which  make  angles  «i,  a2,  <*3,  and  «4,  respectively,  with  the 


FIG.  53. 

horizontal  line  toward  the  right.  The  point  at  which  the  forces 
meet  is  the  free  body.  The  components  along  the  horizontal  are 
PI  cos  ai}  P2  cos  «2,  etc.  If  the  forces  are  in  equilibrium, 

Pi  COS  «i  -f  P2  COS  Q!2  +  P3  COS  0:3  +  P  COS4  a4  =  0.          (1) 

If  the  forces  of  Fig.  53  make  angles  0i,  02,  #3,  and  04  with  a 
second  axis,  the  second  condition  of  equilibrium  is 

Pi  cos  0i  -f  P2  cos  /32  +  P3  cos  03*+  P4  cos  04  =  0.     (2) 

Any  number  of  equations  may  be  written  by  changing  the 
direction  of  the  axis  of  resolution.  Since  it  will  be  shown  later 
that  only  two  such  equations  can  be  independent,  nothing  is 
gained  by  writing  more  than  that  number. 

Example 

A  40-pound  mass  is  supported  by  three  flexible  cords  in  the  same  vertical 
plane.     One  cord  makes  an  angle  of  35  degrees  to  the  left  of » the  vertical 
and  exerts  an  unknown  pull  of  P  pounds.     A  second  cord  makes  an  angle 
of  25  degrees  to  the  right 
of  the  vertical  and  exerts 
an    unknown    pull    of    Q 
pounds.     The  third  cord 
makes  an  angle  of  80  de-. 
grees  to  the  right  of  the 
vertical  and  exerts  a  pull 
of   10   pounds.     Find  the 
unknown   forces   by  reso- 
lutions. 

The  point  0,  Fig.  54,  I, 
at  which   all  these  cords 

meet,  is  the  free  body  in  equilibrium.  Resolving  horizontally  and  consider- 
ing components  toward  the  right  positive  components, 

10  cos  10°  +  Q  sin  25°  -  P  sin  35°  =  0,  (3) 


FIG 


50  MECHANICS  [AKT.  44 

Resolving  vertically  and  considering  upward  components  positive, 

10  sin  10°  +  Q  cos  25°  +  P  cos  35°  =  40.  (4) 

Substituting  the  values  of  the  trigonometric  functions  and  transposing, 

0.4226Q  -  0.5736P  =  -  9.848  (5) 

0.9063Q  +  0.8192P  =  40  -  1.736  =  38.264  (6) 

To  eliminate  Q  multiply  Equation  (5)  by  0.9063  and  Equation  (6)  by  0.4226, 
and  subtract. 

0.4226  X  0.9063Q  -  0.5199P  =  -8.925, 
0 . 4226  X  0 . 9063Q  +  0 . 3462P  =     16 . 170, 
0.8661P  =    25.095, 
P  =  28.97  Ib.  • 

Substituting  the  value  of  P  in  Equation  (5), 

0.4226  Q  =  28.97  X  0.5736  -  9.848  =  16.617  -  9.848  =  6.769. 
Q  =  16.02  Ib. 

This  method  of  horizontal  and  vertical  resolution  involves  two  unknowns 
in  each  equation.  Since  the  coefficients  of  these  unknowns  are  sines  and 
cosines  taken  to  four  significant  figures,  considerable  labor  is  required  in 
order  to  eliminate  one  unknown  and  solve  the  equations.  It  is  better  to 
make  the  resolutions  in  such  a  way  that  one  equation  will  contain  only  one 
unknown.  This  may  be  done  by  resolving  perpendicular  to  the  direction 
of  the  other  unknown  force,  since  the  component  of  a  force  along  a  line 
perpendicular  to  its  direction  is  zero.  In  the  above  example  resolve  along 
the  broken  line  of  Fig.  54,  II,  which  is  perpendicular  to  the  direction  of  Q. 
Since  the  force  Q  makes  an  angle  of  25  degrees  with  the  vertical,  this  line 
perpendicular  to  Q  makes  an  angle  of  25  degrees  with  the  horizontal.  On 
the  right,  this  line  makes  an  angle  of  35  degrees  with  the  direction  of  the 
10-pound  force  and  an  angle  of  65  degrees  with  the  vertical  line  of  the  weight. 
On  the  left  it  makes  an  angle  of  30  degrees  with  the  direction  of  the  force  P. 
Giving  the  positive  sign  to  the  component  toward  the  left,  and  putting  the 
negative  terms  on  the  right  of  the  equality  sign, 

P  cos  30°   =  10  cos  35°  +  40  sin  25°,  (7) 

0.866  P  =  10  X  0.8192  +  40  X  0.4226,  (8) 

0.866  P  =  8.192  +  16.904  =  25.096, 
P   =  28.98. 

The  force  Q  may  now  be  calculated  by  a  resolution  perpendicular  to  the 
force  P,  or  by  substitution  in  Equation  (5).  After  both  P  and  Q  are  found, 
both  may  be  checked  by  substitution  in  Equation  (6), 

LO  X  0.1.736  =  1.736 
16.03  X  0.9063  =  14.528 
28.98  X  0.8192  =  23.740 


40.004  -  40  =  0.004. 


CHAP.  IV]     CONCURRENT  CO-PLANAR  FORCES  51 

Problems 

(Draw  the  space  diagram  for  each  problem.  Mark  each  force  which  acts  on 
the  free  body.) 

1.  A  60-pound  mass  is  suspended  by  two  cords,  one  of  which  makes  an 
angle  of  32  degrees  to  the  left  of  the  vertical,  and  the  other  makes  an  angle 
of  27  degrees  to  the  right  of  the  vertical.     Find  the  tension  in  each  cord 
by  a  resolution  perpendicular  to  the  other  cord,  and  check  by  a  vertical 
resolution.  Ans.     31.78  Ib.  at  32°;    37.09  Ib.  at  27°. 

2.  A  40-pound  mass  is  held  on  a  smooth  inclined  plane,  which  makes  an 
angle  of  27  degrees  with  the  horizontal,  by  means  of  a  cord  parallel  to  the 
plane.     Find  the  tension  in  the  cord  by  a  resolution  parallel  to  the  plane, 
and  find  the  normal  reaction  of  the  plane.     Check  both  by  a  vertical  resolu- 
tion. Ans.     18.16  Ib.  tension;  35.64  Ib.  normal  pressure. 

3.  A  40-pound  mass  is  held  on  a  smooth  inclined  plane,  which  makes  an 
angle  of  27  degrees  with  the  horizontal,  by  means  of  a  cord,  which  makes 
an  angle  of  40  degrees  with  the  horizontal.     Find  the  tension  in  the  cord 
and  the  normal  reaction  and  check. 

Ans.     18.64  Ib.  tension;  31.45  Ib.  normal  pressure. 

4.  A  mass  of  m  pounds  is  held  on  a  smooth  inclined  plane,  which  makes 
an  angle  6  with  the  horizontal,  by  means  of  a  cord  parallel  to  the  plane. 
Find  the  tension  in  the  cord  and  the  normal  reaction  of  the  plane  by  resolu- 
tions parallel   and  perpendicular  to  the  plane,  and  check  by  a  resolution 
vertical.  Ans.  Tension  =  m  sin  6;  normal  =  m  cos  0. 

6.  A  mass  of  m  pounds  is  held  on  a  smooth  inclined  plane,  which  makes 
an  angle  6  with  the  horizontal,  by  means  of  a  horizontal  cord.  Find  the 
tension  in  the  cord  and  the  normal  reaction,  and  check. 

Ans.  Tension  =  m  tan  6;  normal 
"  6.  Figure  55  shows  a  rod  AB,  8  feet 
in  length,  which  is  hinged  at  B  and  sup- 
ported in  a  horizontal  position  by  a  cord  at 
A.  The  cord  is  fastened  to  a  point  C, 
which  is  6  feet  above  the  hinge  B.  A  60- 
pound  mass  is  suspended  from  A.  Neg- 
lecting the  weight  of  the  rod,  and  regard- 
ing all  forces  as  concurrent  at  A,  find  the 
tension  in  the  cord  and  the  horizontal 
reaction  in  the  rod.  Solve  by  vertical 
and  horizontal  resolutions,  and  check  by  a 
resolution  parallel  to  the  cord. 

When  a  force  is  unknown,  both  as  to  direction  and  magnitude, 
it  is  best  to  consider  this  force  as  made  up  of  two  components 
at  right  angles  to  each  other,  and  find  the  magnitude  of  each 
of  these  components  separately.  Next  find  the  direction  and 
magnitude  of  their  resultant,  which  is  the  force  required.  This 
method  is  the  same  as  the  one  for  finding  the  resultant  of  a  set 
of  concurrent  forces. 


52  MECHANICS  [ART.  44 

Example 

A  50-pound  mass  is  supported  by  two  cords,  one  of  which  makes  an  angle 
of  75  degrees  to  the  right  of  the  vertical  and  exerts 
a  pull  of  30  pounds.  Find  the  direction  of  the  second 
cord  and  the  tension  which  it  exerts. 

In  Fig.  56,  the  unknown  force  Q  at  an  unknown 
angle  6  with  the  vertical  may  be  regarded  as  made 
up  of  a  horizontal  component  H  and  a  vertical 
component  V.  Resolving  horizontally, 

H  =  30  cos  15°  =  30  X  0.9659  =  28.977. 
FIG.  56.  Resolving  vertically, 

V  =  50  -  30  sin  15°  =  50  -  30  X  0.2588  =  50  -  7.764  =  42.236, 


6  =  34°  27', 

_  42.236  _  42.236  _ 
y  "  ~^oiY      08246 
Check  by  resolving  parallel  to  Q, 

50  cos  34°  27'  +  30  sin  19°  27'  =  Q, 
50  X  0.8246  =  41.23 
30  X  0.3330  =    9.99 

51.22  =  Q 


Problems 

7.  A  mass  of  80  pounds  is  supported  by  three  cords  in  the  same  vertical 
plane.     One  cord  exerts  a  pull  of  20  pounds  in  a  horizontal  direction  toward 
the  right.     The  second  cord  makes  an  angle  of  35  degrees  to  the  right  of  the 
vertical  and  exerts  a  pull  of  50  pounds.     Find  the  direction  of  the  third 
cord  and  the  tension  which  it  exerts.     Check. 

8.  A  50-pound  mass  is  supported  by  two  cords,  one  of  which  makes  an 
angle  of  38  degrees  to  the  right  of  the  vertical,  and  the  other  exerts  a  pull 
of  35  pounds.     Find  the  tension  in  the  first  cord,  and  the  direction  of  the 
second.     Check  the  results. 

Ans.     23°  35'  or  80b  25'  with  the  vertical;  22.74  Ib.  or  56.06  Ib. 

9.  A  mass  of  100  pounds  is  supported  by  two  cords,  each  of  which  makes 
an  angle  of  55  degrees  with  the  vertical.     Find  the  tension  in  each  by  two 
resolutions.     Check. 

10.  Solve  Problem  9  when  each  cord  makes  an  angle  of  80  degrees  with 
the  vertical.     Check  by  means  of  the  force  triangle  drawn  to  the  scale  of 
1  inch  =  40  pounds. 

11.  A  mass  of  80  pounds  is  supported  by  two  cords,  one  of  which  exerts 
a  pull  of  60  pounds;  the  other  exerts  a  pull  of  50  pounds.     Find  the  direction 
of  each  cord  by  horizontal  and  vertical  resolutions. 

Ans.     The  cord  which  pulls  60  pounds  makes  an  angle  of  38°  38'  with  the 
vertical.    The  other  cord  makes  an  angle  of  48°  31'  with  the  vertical. 


CHAP.  IV]     CONCURRENT  CO-PLANAR  FORCES 


53 


45.  Trigonometric  Solution. — When  three  concurrent  forces 
are  in  equilibrium,  the  force  polygon  is  a  triangle.  It  is  some- 
times convenient  to  draw  this  triangle  and  calculate  the  un- 
known forces  or  directions  trigonometrically.  This  method  is 
especially  desirable  when  two  of  the  forces  are  perpendicular 
to  each  other  so  that  the  force  polygon  is  a  right-angled  triangle. 
It  is  the  best  method  when  the  forces  are  expressed  literally  in- 
stead of  numerically.  When  the  forces  form  an  oblique-angled 
triangle,  and  the  known  forces  are  given  in  numbers,  the  method 
of  resolutions  is  better. 

Example 

A  40-pound  mass  is  supported  by  two  cords,  one  of  which  is  horizontal 
and  exerts  a  pull  of  16  pounds.  Find  the  direction  of  the  second  cord  and 
the  tension  which  it  exerts. 

The  force  triangle,  Fig.  57,  is  a  right- 
angled  triangle  of  base  16  and  altitude 
40.  Tan  0  =  0.4,  P  =  40  sec  6. 

Problems 

(In  these  problems  draw  the  space 
diagram  and  the  force  diagram.  Unless 
it  is  desired  to  check  by  means  of  the 
graphical  solution,  it  is  not  necessary 
that  these  diagrams  be  drawn  to  scale. 
The  force  diagram  must  be  drawn  sepa- 
rate from  the  space  diagram. ) 

1.  Selve  Problem  2  of  Art.  44  by  the  trigonometric  solution  of  the  force 
triangle. 

2.  Solve  Problem  6  of  Art.  44  by  means  of  the  force  triangle. 

3.  Solve  Problem   1  of  Art.  44  by  means  of  the  oblique-angled  force 
triangle.     Use  the  law  of  sines.     Is  the  method  shorter  than  that  of  two 
resolutions? 

/  4.  A  mass  of  m  pounds  is  held  on  a  smooth  inclined  plane,  which  makes 
an  angle  a  with  the  horizontal,  by  a  rope  which  makes  an  angle  ft  with  the 
horizontal.  Find  the  tension  in  the  rope  and  the  normal  reaction  by  means 
of  the  solution  of  the  force  triangle. 


40  I  b. 


16  Ib. 
FIG.  57. 


Ans.     Tension  =  _ -==r- 


cos  (a  —  /3) 


;  normal 


m  cos 


cos  (a  —  j8) 


It  sometimes  happens  that  the  force  triangle  is  similar  to  a 
triangle  of  the  space  diagram.  In  that  case  the  magnitudes  of  the 
forces  are  proportional  to  the  corresponding  lengths  in  the  space 
diagram. 


54 


MECHANICS 
Example 


[ART.  46 


Figure  58  shows  a  horizontal  rod,  12  feet  in  length,  which  is  hinged  at  the 
left  end  A.  A  cord  is  attached  to  the  right  end  B  and  fastened  to  a  point  C, 
which  is  8  feet  above  A.  Neglecting  the  weight  of  the  rod,  find  the  ten- 
sion in  the  cord  and  the  compression  in  the  rod  when  a  load  of  40  pounds 
is  placed  at  B. 


The  force  of  40  pounds  in  the  force  diagram  is  homologous  to  the  length 
of  8  feet  in  the  space  diagram.  The  ratio  is  5  to  1.  The  horizontal  push 
in  the  rod  is  5  X  12  =  60  pounds.  In  the  space  diagram  the  length  BC  is 
A/208  =  14.422  feet.  The  tension  in  the  cord  is  5  X  14.422  =72.11  pounds. 

Problems 

6.  In  Fig.  58,  the  cord  BC  is  shortened  until  the  point  B  is  2  feet  higher 
than  A.  Find  the  compression  in  the  bar  and  the  tension  in  the  cord. 

Ans.     60  lb.;  66.33  Ib. 

6.  If  the  length  AB  of  Fig.  58  is  constant  and  the  point  C  is  at  a  constant 
distance  directly  above  A,  show  that  the  compression  in  the  rod  will  be  the 
same  for  any  position  of  AB. 

46.  Number  of  Unknowns. — In  each  of  the  problems  of  the 
equilibrium  of  concurrent,  coplanar  forces  in  the  preceding 
articles  there  have  been  two  unknown  quantities.  These  have 
been: 

(1)  An  unknown  magnitude  and  an  unknown  direction  of  one 
force; 

(2)  An  unknown  magnitude  of  one  force  and  an  unknown 
direction  of  another; 

(3)  Two  unknown  magnitudes; 

(4)  Two  unknown  directions. 

With  two  unknowns,  two  independent  equations  are  required  to 
solve  each  problem. 

As  far  as  the  mechanics  of  the  problem  is  concerned,  there  may 
be  only  two  unknowns  in  a  problem  of  concurrent,  coplanar 


CHAP.  IV]      CONCURRENT  CO-PLANAR  FORCES  55 

forces  in  equilibrium.  This  may  be  shown  in  several  ways.  If 
the  problem  is  solved  by  resolutions,  only  two  equations  can  be 
written  which  are  independent  of  each  other.  Fig.  59  shows  a 
force  AB.  Its  orthographic  component  along  a  direction  OC  is 
the  length  A\Bi.  Its  orthographic  component  along  another 
direction  OD  is  the  length  A2B2.  If  a  third  orthographic  com- 
ponent were  drawn,  its  length  could  be  expressed  in  terms  of 
AiBi,  A2B2,  and  the  angles.  Consequently, 
if  three  equations  were  written  for  a  problem 
of  concurrent,  coplanar  forces,  any  one  of 
these  equations  could  be  derived  algebraically 
from  the  other  two,  and  would  not  be  inde- 
pendent. Since  there  can  be  only  two  inde- 
pendent equations  for  a  problem  of  concurrent,  °  *•  B, 
coplanar  forces,  it  is  evident  that  there  may 
be  only  two  unknowns,  unless  conditions  are  given  which  are 
not  based  upon  mechanics. 

The  number  of  possible  unknowns  may  be  determined  in  a 
different  way  from  the  geometry  of  the  force  polygon.  All  the 
known  forces  may  be  represented  in  the  force  diagram  by  the 
single  vector  of  their  resultant.  The  problem  of  equilibrium 
involves  the  solution  of  a  polygon  of  which  this  resultant  vector 
is  one  side.  This  polygon  is  a  triangle  for  all  problems  which 
may  be  solved  from  mechanical  considerations.  A  triangle  may 
be  solved  in  the  following  cases: 

1.  One  side  and  two  angles  given,  with  the  length  of  two  sides 
unknown. 

2.  Two  sides  and  the  included  angle  given,  with  the  magnitude 
and  direction  of  the  third  side  unknown. 

3.  One  side  given  together  with  the  length  of  a  second  side 
and  the  direction  of  the  third  side.     The  length  of  one  side  and 
the  direction  of  another  are  unknown. 

4.  The  length  of  all  sides  given,  with  two  directions  unknown. 
Some  of  these  cases  apparently  have  three  unknowns.      When 
two  angles  of  a  triangle  are  given,  the  third  angle  is  regarded 
as  known. 

It  is  evident  from  the  foregoing  statements  that  a  problem  of 
the  equilibrium  of  concurrent,  coplanar  forces  may  have  only 
two  unknown  quantities.  Sometimes  it  happens  that  there  is  a 
greater  number  of  unknowns.  Such  a  problem  can  not  be  solved 
completely  unless  additional  conditions  are  given. 


56 


MECHANICS 


[ART.  47 


Example 

A  body  is  in  equilibrium  under  the  action  of  three  forces.  These  are: 
a  force  of  40  pounds  horizontally  toward  the  right,  a  force  P  at  ^n  angle  of 
25  degrees  to  the  left  of  the  vertical,  and  a  force  Q  of  unknown  direction 
and  magnitude. 

There  are  three  unknowns,  and  an  attempt  at  a  graphical  solution,  Fig. 
60,  shows  that  it  may  be  solved  in  an  infinite  number  of  ways.  Another 
condition  must  be  added  to  make  the  problem  definite.  Let  the  force  Q 
of  unknown  direction  be  twice  as  great  as  the  force  P  at  25  degrees  to  the  left 
of  the  vertical.  This  is  an  additional  algebraic  condition  which  does  not 
depend  upon  mechanics. 

Q  =  2P  (1) 

Resolving  horizontally  and  vertically, 

P  sip  25°  +  2P  sin  0  =  40,  (2) 

P  cos  25°  =  2P  cos  8,  (3) 

cos  25°      0.9063 


cos  e 


0.4531. 


Problems 

1.  In  Fig.  60,  let  the  force  Q  be  three  times  the  force  P.     Find  the  mag- 
nitude of  each  force  and  the  direction  of  Q. 

Ans.     P  =  12.19  lb.;  Q  =  36.57  Ib. 

2.  Solve  Problem  1  graphically. 

3.  A  body  is  in   equilibrium   under 
the  action  of  four  forces.     These  are: 
a  force   of  40  pounds  at  0  degrees,  a 
force  Q  at  110  degrees,  a  force  P  at  245 
degrees,  and  a  force  equal  to  Q  at  180 
degrees.     Solve  graphically. 

47.  Moment  of  a  Force. — Figure  61  shows  a  bar  OB,  which  is 
hinged  at  0  and  has  a  load  P  applied  at  B.  The  load  tends  to 
turn  the  bar  about  the  hinge, 
and  does  turn  it  unless  it  is 
balanced  by  a  second  force 
tending  to  turn  in  the  oppo- 
site direction.  In  Fig.  61, 
I,  the  turning  effect  of  the 
force  is  greater  than  in  the 
position  of  Fig.  61,  II.  In 
the  position  of  Fig.  61,  III, 
the  force  has  no  tendency  to 
turn  the  bar  about  the  hinge. 
In  the  other  positions,  the  force  P  is  said  to  exert  a  moment 
on  the  bar  about  the  hinge.  The  point  about  which  moment 


III 


€> 


FIG.  61. 


CHAP.  IV]     CONCURRENT  CO-PLANAR  FORCES 

is    taken   is    called    the    origin   of    moments   or   the    center   of 
moments.          r 

Definition^-The  moment  of  a  force  about  a  point  in  its  plane  is 
the  product  of  the  magnitude  of  the  force  multiplied  by  the  perpendicu- 
lar distance  from  the  point  to  the  line  along  which  the  force  acts. 

In  Fig.  61,  the  moment  of  the  force  P  is  the  product  of  the 
force  multiplied  by  the  length  OC.     If  the  force  is  measured  in 
pounds  and  the  distance  in  feet,  the  moment  is  expressed   in 
foot-pounds.     To  distinguish  moment  from  work,  some  writers        ^ 
use  pound-feet  for  moment  and  foot-pounds  IO^W-I^H 

for  work.     Though  this  distinction  is  desirable,  Pi    ee\f*tfa&*  « 

it  has  not  come  into  general  use.  *&'''   ^~-  *  — 

The  perpendicular  distance  from  the  origin  tfs&B      -e.-  >o^J* 

of  moments  to  the  line  of  action  of  the  force  is 

the  effectiyejrioment  arm.  The  length  OB,  from  •o^.crs-n.&..>(  fi^^-Py 
the  origin  of  moments  to  the  point  of  appli- 
cation of  the  force,  may  be  called  the  apparent 
moment  arm.  If  the  length  of  the  apparent  mo- 
ment arm  is  a,  Fig.  62,  and  the  angle  between 
its  direction  and  the  direction  of  the  force  is  a, 

effective  arm  =  a  sin  a  (1)  FlG   62 

moment  =  P  X  a  sin  a.         Formula  IV 

Counter-clockwise  moment  is  generally  regarded  as  positive, 
especially  in  works  of  mathematical  nature;  and  clockwise 
moment  is  regarded  as  negative.  In  algebraic  equations,  mo- 
ment will  be  represented  in  this  book,  by  M. 

Problems 

1.  A  force  of  60  pounds,  at  an  angle  of  25  degrees  to  the  right  of  the  ver- 
tical, is  applied  to  the  right  end  of  a  bar,  which  is  3  feet  in  length.     The  bar 
makes  an  angle  of  80  degrees  to  the  right  of  the  vertical.     Find  the  moment 
of  this  force  about  the  left  end  of  the  bar. 

Ans.     M  =  147.45  foot-pounds. 

2.  A  force  of  12  pounds,  at  an  angle  of  20  degrees  to  the  left  of  the  vertical, 
is  applied  to  a  point  whose  coordinates  are  (4,  3).     Find  the  moment  of 
this  force  about  the  origin  of  cobrdinates. 

Ans.     M  =  57.42  foot-pounds. 

3.  Solve  Problem  2  if  the  force  is  at  an  angle  of  20  degrees  to  the  right  of 
the  vertical. 

By  changing  the  order  of  the  letters  of  Formula  IV,  it  may  be 
written, 

M  =  P  sin  a  X  a.  (2) 


58  'MECHANICS  [ART.  48 

The  term  P  sin  a  is  the  component  of  the  force  perpendicular  to 
the  direction  of  the  apparent  moment  arm.  Equation  (2)  gives 
this  second  definition  of  moment :  The  moment  of  a  force  is  the 
product  of  the  apparent  arm  multiplied  by  the  component  of  the 
force  which  is  perpendicular  to  it. 

The  first  definition  states  that  moment  is  the  entire  force 
multiplied  by  the  component  of  the  arm;  the  second  definition 
states  that  moment  is  the  product  of  the  entire  arm  multiplied 
by  the  component  of  the  force. 

Since  force  and  apparent  arm  are  vectors,  moment  is  the  prod- 
uct of  two  vectors.  It  will  be  shown  in  Art.  108  that  this  product 
is  a  vector. 

48.  Moment  of  Resultant. — If  PI,  P2,  PS,  etc.  are  forces  which 
make  angles  ai,  a2,  «s,  etc.  with  the  apparent  arm  of  length  a, 
the  sum  of  the  moments  of  all  these  forces  about  the  origin  is 

M  =  PI  sin  aia  +  P2  sin  aza  +  P3  sin  a3a  =  (Pi  sin  «i  + 
P2  sin  a  2  +  PS  sin  a3)a.  The  term  PI  sin  «i  +  P2  sin  «2  +  ' 
P3  sin  as  +  etc.  in  the  above  equation  is  the  component  of  the 
resultant  force  perpendicular  to  the  apparent  arm;  conse- 
quently, the  moment  of  the  resultant  of  several  concurrent  forces  is 
equal  to  the  sum  of  the  moments  of  separate  forces. 

When  the  moment  of  a  force  is  calculated  by  means  of  the  com- 
ponent perpendicular  to  the  apparent  arm,  it  is  really  calculated 
by  means  of  the  moment  of  two  components  at  right  angles  to 
each  other.  From  Fig.  62,  II,  the  second  component  is  P  cos  a 
in  the  direction  of  the  apparent  arm.  Since  the  line  of  action  of 
this  component  passes  through  the  origin  of  moments,  its  mom- 
ent is  zero.  The  total  moment  is,  then,  the  moment  of  the  com- 
ponent normal  to  the  apparent  arm. 

It  is  often  advisable  to  resolve  both  the  force  and  the  apparent 
arm  into  components.  This  method  is  especially  convenient 
when  the  coordinates  of  the  ends  of  the  apparent  arm  are  given 
instead  of  its  length  and  direction. 

Example 

A  force  of  40  pounds  at  an  angle  of  25  degrees  to  the  left  of  the  vertical 
is  applied  at  the  point  whose  coordinates  are  (6,  2).  Find  the  moment  of 
this  force  about  the  origin  of  coordinates. 


CHAP.  IV]     CONCURRENT  CO-PLANAR  FORCES  59 

First  method. 

The  apparent  moment  arm  is  OB,  Fig.  63. 
OB  =  6.325  ft. 
Tan   6  =  *  ;  6    =18°    26'. 

OC  =  OB  cos  (25°  -  9)  =  OB  cos  6°  34'  =  6,325  X  0.9934  =  6.283  ft. 
M  =  40  X  6.283  =  251.32  foot-pounds. 


40 


Second  method. 
Resolving  horizontally,  Fig.  63,  II, 

H  =  40  X  0.4226  =  16.904. 
Resolving  vertically, 

V  =  40  X  0.9063  =  36.252.    >    \ 

The  effective  arm  of  the  horizontal  component  is  the  vertical  length  of  2 
feet.  The  effective  arm  of  the  vertical  component  is  the  horizontal  distance 
of  6  feet.  Both  components  turn  counter-clockwise. 

16.904  X  2  =     33.808 
36. 252  X  6  =  217.512 

25 1.320  foot-pounds. 


Problems 

1.  Solve  the  example  above  if  the  force  makes  an  angle  of  25  degrees  to 
the  right  of  the  vertical. 

2.  A  force  of  50  pounds  at  an  angle  of  15  degrees  to  the  right  of  the  vertical 
is  applied  at  the  point  (5,  4).     Find  the  moment  of  this  force  about  the 
origin  of  coordinates,  and  also  about  the  point  (2,  2). 

A  force  may  be  regarded  as  applied  at  any  point  along  its 
line  of  action.  The  effective  moment  arm  in  Fig.  63,  III,  is  the 
same,  whether  the  force  is  applied  at  B  or  D,  or  at  any  other 
point  along  its  line.  If  the  force  of  40  pounds  be  regarded  as 
applied  at  D,  on  the  axis  of  X,  the  moment  about  0  is  the  product 
of  the  length  OD  multiplied  by  the  vertical  component  of  the 
force. 


60  MECHANICS  [ART.  49 

Problems 

3.  Calculate  the  length  OD  of  Fig.  63,  III,  trigonometrically,  and  multi- 
ply by  the  vertical  component  of  the  40-pound  force. 

4.  A  force  of  50  pounds  at  an  angle  of  40  degrees  to  the  left  of  the  vertical 
passes  through  the  point   (4,  4).     Find  its  moment  about  the  origin  of 
coordinates.     Solve  by  both  methods  of  the  example  above.     Also  find  the 
intersection  of  its  line  of  action  with  the  X  axis,  and  multiply  the  abscissa 
of  this  point  by  the  vertical  component  of  the  force.     Find  the  intersection 
with  the  y  axis  and  multiply  the  F-intercept  by  the  horizontal  component. 

5.  A  force  of  10  pounds  at  an  angle  of  35  degrees  to  the  left  of  the  vertical 
passes  through  the  point  (6,  5).     Find  its  moment  with  respect  to  the  point 
(2,  2).     Solve  by  means  of  the  horizontal  and  vertical  components.     Solve 
by  the  vertical  component  alone.     Solve  by  the  horizontal  component  alone. 
Find  the  distance  of  the  point  (2,  2)  from  the  line  of  action  of  the  force. 

6.  Find  the  distance  of  the  point  (3,  5)  from  a  line  through  the  point 
(6,  7)  at  an  angle  of  20  degrees  to  the  left  of  the  vertical.     Solve  by  princi- 
ples of  moments. 

7.  A  force  of  20  pounds  is  applied  along  a  line  which  passes  through  the 
points  (0,  6)  and  (8,  0).     Find  its  moment  about  the  origin  of  coordinates. 
Solve  by  the  vertical  component  with  the  horizontal  arm.     Check  by  the 
horizontal  component  with  the  vertical  arm. 

49.  Equilibrium  by  Moments. — Since  the  moment  of  the 
resultant  of  a  set  of  concurrent  forces  is  equal  to  the  sum  of  the 
moments  of  the  separate  forces,  and,  since  the  resultant  of  a 
set  of  forces  in  equilibrium  is  zero,  it  follows,  that  the  sum  of  the 
moments  of  a  set  of  concurrent  forces  about  any  origin  is  zero  when 
the  forces  are  in  equilibrium.  This  affords  another  method  of 
solving  a  problem  of  the  equilibrium  of  concurrent,  coplanar 
forces.  The  moment  equation  of  equilibrium  is 

M  =  PI  sin  ai  a  +  P2  sin  a2  a-  +  PS  sin  a3  a  +  etc.  =0'.   (1) 
M  —  (Pi  sin  ai  +  P%  sin  a2  +  PS  sin  a3  +  etc.)  a  =  0.      (2) 

The  term  in  the  parenthesis  in  Equation  (2)  is  the  sum  of  the 
components,  perpendicular  to  the  apparent  moment  arm  of 
length  a,  of  all  the  forces  which  act  on  the  free  body.  It  is 
evident  from  this  equation,  after  it  has  been  divided  by  the 
apparent  arm,  that  a  moment  equation  for  concurrent  forces  is 
equivalent  to  a  resdhitipn  perpendicular  to  the  apparent  arm. 

In  the  solution  of  a  problem  of  concurrent  forces,  a  moment 
equation  is  used,  instead  of  a  resolution  equation,  when  it  is 
easier  to  calculate  the  length  of  the  effective  arm  than  it  is  to 
resolve  the  force  into  its  components.  In  a  structure  or  machine 


CHAP.  IV]     CONCURRENT  CO-PLANAR  FORCES 


61 


FIG.  64. 


the  effective  force  may  often  be  measured  directly.  Figure  64 
represents  a  bar  hinged  at  0  and  supporting  a  load  P  at  B.  The 
bar  is  supported  by  a  cord  at  B.  It  is  desired  to  find  the  tension 
in  this  cord.  The  three  forces  at  B  are  the  load  P,  the  tension  in 
the  cord  T,  and  the  compression  along  the  bar.  If  moments  are 
taken  about  the  hinge  0,  the  line  of  action  of  the  compression  in 
the  bar  passes  through  this  point  so  that  its  moment  is  zero. 
The  moment  of  T  in  a  counter-clock- 
wise direction  must  be  equal  to  the 
moment  of  P  in  a  clockwise  direc- 
tion. The  effective  moment  arm  of 
T  is  the  length  OD,  measured  from  0 
perpendicular  to  the  direction  of  the 
cord.  The  moment  arm  of  P  is  the 
perpendicular  distance  from  the  hinge 
to  its  line  of  action — the  length  OC  of 
Fig.  64.  To  find  the  perpendicular 
distance  from  a  point  to  a  straight 
line,  it  is  only  necessary  to  measure 
the  shortest  distance.  This  may  be 
done  with  a  tape-line  or  rule.  In  the  actual  machine  or  struc- 
ture, it  is  much  easier  to  measure  these  lengths  than  it  is  to 
determine  the  angles  necessary  for  resolution  equations. 
The  moment  equation  for  Fig.  64  is 

T  X  OD  =  P  X  OC  (3) 

To  find  the  compression  in  the  bar  of  Fig.  64,  moments  may  be 
taken  about  some  point  in  the  line  of  the  cord.  The  point  G 
may  well  be  used  as  the  center  of  moments. 

Problems 

1.  A  mass  of  200  pounds  is  suspended  by  two  cords,  which  are  fastened 
to  the  mass  at  a  point  A.     One  cord  is  attached  to  a  fixed  point,  which  is 
5  feet  higher  than  A  and  8  feet  to  the  left  of  the  vertical  line  through  it. 
The  second  cord  is  attached  to  another  point,  which  is  7  feet  higher  than 
A  and  4  feet  to  the  right  of  the  vertical  line  through  it.     Draw  the  space 
diagram  to  scale.     Find  the  tension  in  the  left  cord  by  moments  about  some 
point  in  the  line  of  the  right  cord,  measuring  the  moment  arms  from  the 
drawing.     Then  find  the  tension  in  the  right  cord  by  moments  about  some 
point  in  the  line  of  the  left  cord.     Check  by  moments  about  a  point  directly 
over  the  load. 

2.  A  40-pound  mass  is  supported  by  two  cords,  one  of  which  is  horizontal 
and  the  other  makes  an  angle  of  23  degrees  with  the  vertical.     Find  the 


62  MECHANICS  [ART.  50 

tension  in  the  horizontal  cord  by  moments  about  some  point  in  the  line  of 
the  other  cord,  computing  the  effective  arm  trigonometrically.  Find  the 
tension  in  the  other  cord  by  resolutions. 

3.  In  an  arrangement  similar  to  Fig.  64  the  bar  is  weightless,  is  6  feet 
long,  and  makes  an  angle  of  25  degrees  with  the  horizontal.     The  cord  is 
attached  to  a  fixed  point  E  which  is  5  feet  above  the  hinge  O.     Draw  the 
space  diagram  to  the  scale  of  1  inch  =  1  foot.     With  the  hinge  as  the  origin 
of  moments,  measure  the  effective  arms  on  the  space  diagram  and  calculate 
T  when  the  load  P  is  60  pounds.     Find  the  compression  in  the  bar  with 
the  origin  of  moments  at  G,  at  a  distance  of  1  foot  from  E.     Also  solve  for 
the  compression  with  the  origin  of  moments  at  E. 

4.  Solve  Problem  3  by  moments,  calculating  the  arms  instead  of  measuring 
them. 

5.  Solve  Problem  6  of  Art.  44  by  moments. 

50.  Conditions  for  Independent  Equations. — In  a  problem  of 
concurrent,  coplanar  forces  there  are  two  independent  equations. 
These  may  be : 

(1)  Two  resolution  equations; 

(2)  Two  moment  equations; 

(3)  One  resolution  equation  and  one  moment  equation. 

A  moment  equation  is  equivalent  to  a  resolution  perpendicular 
to  the  line  joining  the  origin  of  moments  to  the  point  of  applica- 
tion of  the  forces.  If  two  moment  equations  are  written,  and 
the  line  joining  the  point  of  application  of  the  forces  with  one 
origin  of  moments  passes  through  the  other  origin  of  moments, 
the  equations  will  not  be  independent.  Each  moment  equation 
will  be  equivalent  to  a  resolution  perpendicular  to  this  line. 
After  the  equations  have  been  divided  by  the  lengths  of  the 
apparent  moment  arms,  they  are  identical.  When  two 
moment  equations  are  written  for  a  problem  of  concurrent,  coplanar 
forces,  the  two  origins  of  moment  and  the  point  of  application  of  the 
forces  must  not  lie  in  the  same  straight  line.  When  one  resolution 
and  one  moment  equation  are  written,  the  resolution  must  not  be 
perpendicular  to  the  line  joining  the  origin  of  moments  with  the 
point  of  application  of  the  forces. 

A  resolution  perpendicular  to  the  direction  of  an  unknown  force 
eliminates  that  force.  A  moment  about  a  point  in  the  line  of 
action  of  an  unknown  force  eliminates  that  force. 

61.  Connected  Bodies. — It  frequently  happens  that  two  points, 
each  of  which  is  subjected  to  a  set  of  concurrent  forces,  are  con- 
nected together  by  a  single  member,  such  as  a  cord  or  hinged  rod. 


CHAP.  IV]     CONCURRENT  CO-PLANAR  FORCES 


63 


40' 


FIG.  65. 


Each  point  is  in  equilibrium  under  the  action  of  the  external 
forces  and  of  the  force  in  the  connecting  member. 

In  Fig.  65,  the  forces  are  concurrent  at  A  and  B.  The  member 
which  joins  A  to  B  exerts  a  downward  pull  T  on  B  and  an  equal 
upward  pull  on  A.  Including  the 
direction  and  magnitude  of  the  un- 
known tension  T,  there  are  four 
unknowns  at  B.  It  is  not  possible 
to  solve  for  these  unknowns  by  equi- 
librium equations  at  B,  unless  the 
number  of  unknowns  is  reduced  to 
two.  At  A,  on  the  other  hand,  there 
are  only  two  unknowns,  and  the 
problem  may  be  solved.  In  prob- 
lems of  this  kind,  begin  at  a  point 
where  there  are  only  two  unknowns. 

The    equilibrium    at    A    may    be 
solved    by   one    moment    and    one 
resolution   equation.     If  the  length 
A  B  is  a  and  the  angle  with  the  vertical  is  6,  moments  about  B 
give, 

10  a  cos  6  =  40  a  sin  0,  (1) 

tan0   =  0.25;  0  =  14°  02'. 

A  vertical  resolution  gives, 

T7  cos  0  =  40;  T  =  40  sec  0  =  40  X  1.0307  =  41.  23.  Ib. 

The  force  T  may  be  checked  from  the  force  triangle  by  means  of 
the  square  root  of  the  sum  of  the  squares  of  10  and  40.  This 
method  is  convenient  when  the  numbers  are  small.  The  trigo- 
nometric method  is  better  for  large  numbers. 

A  closed  curve  has  been  drawn  about  the  point  A  of  Fig.  65. 
This  curve  encloses  the  joint,  which  is  the  free  body.  The 
arrows  inside  the  curve  represent  the  direction  of  the  forces 
which  act  on  the  free  body.  The  arrow  in  the  member  AB  points 
upward,  which  shows  that  AB  pulls  upward  on  the  joint.  Since 
AB  pulls  upward  on  the  joint  at  A,  the  joint  pulls  downward  on 
A  B.  The  stress  in  AB  is  tensile. 

At  joint  B  the  tension  in  AB  pulls  downward.  The  direction 
and  magnitude  of  this  tension  are  known.  Its  components  are 
10  pounds  horizontally  toward  the  left  and  40  pounds  vertically 
downward,  The  force  T  at  A  is  the  equilibrant  of  the  forces  of 


64 


MECHANICS 


(ART.  51 


10  pounds  and  40  pounds.  The  force  T  at  B  is  the  resultant  of 
these  two  forces.  The  unknowns  at  B  are  now  the  forces  P  and 
Q  of  known  direction.  The  free  body  is  now  the  joint  B.  Re- 
solving perpendicular  to  Q  and  using  the  components  of  T 
instead  of  the  resultant  force, 

P  sin  70°  +  10  cos  40°  =  40  sin  40°  (2) 

40  X  0.6428  =    25.712 

- 10  X  0.7660  =  -7.660 

0.9397  P  =    18.052 

P  =    19.20  Ib. 
Resolving  vertically, 

Q  cos  40°  +  P  cos  30°  =  40  (3) 

0.7660Q  =  40  -  19.20  X  0.8661  =  40  -  16.64  =  23.36 
Q  =  30.49  Ib. 

These  results  may  be  checked  by   horizontal  resolution. 

Problems 

1.  In  Fig.  66,  find  the  unknown  forces  at  the  right  end  by  one  moment 
and  one  resolution  equation  or  by  two  resolution  equations.  Put  the  arrows 
showing  the  direction  of  the  forces  R  and  Q  inside  the  closed  curve. 

Ans.     R  =  160  Ib.  compression;  Q  =  200  Ib.  tension. 


2.  "Using  the  value  of  Q  found  in  Problem  1  (or  its  components)  find  the 
magnitude  of  the  forces  P  and  T  at  the  top.  Draw  the  arrows  inside  the 
closed  curve  at  this  point  to  show  the  forces  acting  on  the  joint. 

Ans.  T  =  241.7  Ib.  tension;  P  =  166.9  Ib.  tension;  horizontal  com- 
ponent of  P  =  1 18.02  Ib. 


CHAP.  IV]     CONCURRENT  CO-PLANAR  FORCES 


65 


3.  At  the  left  end,  put  arrows  in  the  closed  curve  to  give  the  direction  of 
the  forces  P  and  R.     Find  the  horizontal  and  vertical  components  of  the 
hinge  reaction  S,  and  find  the  direction  and  magnitude  of  S. 

Ans.     H  =  41.98  Ib.  toward  the  right;  V  =  118.02  Ib.  downward. 
S  =  125.3  Ib.  at  19°  35'  with  the  vertical. 

4.  Solve  Problems  1,  2,  and  3  if  the  force  T  is  vertical. 

5.  Solve  Problems  1,  2,  and  3  if  the  force  T  makes  an  angle  of  10  degrees 
to  the  right  of  the  vertical. 


52.  Bow's  Method. — The  forces  in  a  connected  system  are 
often  determined  graphically.  It  is  possible  to  begin  at  some 
point  at  which  there  are  only  two  unknowns  and  draw  the  force 
polygon.  The  force  in  the  member  which  joins  this  point  with 
a  second  point  may  then  be  used  in  the  force  polygon  for  that 
point.  Instead  of  drawing  two  lines  for  the  force  in  the  connect- 
ing member,  and  making  two  separate  force  polygons,  it  is 
better  to  draw  this  common  line  but  once,  and  to  connect  the 
two  polygons  into  a  single  diagram.  To  avoid  confusion  in 
complicated  diagrams,  a  system  of  lettering  the  space  diagram 
and  force  diagram  has  been  adopted  by  many  workers  in  graphi- 
cal statics.  This  is  called,  from  the  name  of  the  inventor,  Bow's 
method.  Each  letter  in  the  space  diagram  represents  an  area 
bounded  by  members  which  transmit  force  in  the  direction  of 
their  length.  The  force  in  a 
given  member  is  represented 
by  the  two  letters  which  the 
member  separates  in  the  space 
diagram.  In  the  force  dia- 
gram, one  of  these  letters  is  put 
at  each  end  of  the  line  repre- 
senting the  force.  In  this 
book,  Italic  capitals  will  be 
used  on  the  space  diagram  and 
lower  case  Italics  on  the  force 
diagram. 

In  Fig.  67,  the  letter  A,  on 
the  space  diagram,  represents  the  area  to  the  left  of  the  vertical 
cord  and  below  the  horizontal  cord.  The  letter  B  represents  the 
area  above  the  horizontal  cord  and  to  the  left  of  the  cord  at 
30  degrees  with  the  vertical.  The  horizontal  cord  in  the  space 
diagram  separates  the  area  A  from  the  area  B.  The  line  from  a  to 
b  in  the  force  diagram  represents  the  force  in  the  horizontal  cord. 

5 


40 


Ib. 


Force  Diagram 
Space  Diagram 

FIG.  67. 


66 


MECHANICS 


[ART.  52 


To  solve  the  problem  graphically,  the  space  diagram  is  first 
drawn  to  scale  and  the  spaces  are  lettered.  The  direction  and 
magnitude  of  the  force  in  the  vertical  cord  are  known.  To 
represent  the  direction,  an  arrow  pointing  downward  is  placed 
on  this  member  in  the  space  diagram.  The  force  diagram  is 
begun  by  drawing  a  vertical  line  40  units  in  length.  Since  the 
vertical  cord  separates  area  A  from  area  Z),  one  end  of  this  line 
in  the  force  diagram  is  marked  a  and  the  other  end  is  marked  d. 
In  Fig.  67,  the  letter  a  is  put  at  the  top.  It  would  be  just  as 
correct  to  put  d  at  the  top.  An  arrow  is  put  in  the  force  diagram 
to  show  the  direction  of  the  force  in  ad.  Since  the  horizontal 
cord  separates  area  A  from  area  B,  the  force  in  this  member  is  ab. 
A  line  of  indefinite  length  is  drawn  horizontally  through  a  in  the 
force  diagram.  The  third  force  at  this  joint  is  bd.  Through  the 
point  d  of  the  force  diagram,  a  line  is  now  drawn  parallel  to  the 
cord  which  separates  B  from  D  in  the  space  diagram.  The  inter- 
section of  this  line  through  d  with  the  horizontal  line  through  a 
gives  the  point  b.  The  arrows  are  now  placed  in  the  force  dia- 
gram. Since  the  arrow  in  ad  points  toward  d  the  arrow  in  db 
must  point  from  d  toward  b  and  the  arrow  in  ba  from  b  toward  a. 
Corresponding  arrows  are  now  drawn  at  the  lower  joint  in  the 
These  fe"  shown  in  Fig.  68. 

Figure  68  shows  the  final 
diagrams  for  both  joints, 
while  Fig.  67  gives  the  first 
steps  in  the  construction  of 
these  diagrams.  To  avoid 
confusion,  the  joints  are 
numbered  1  and  2.  The 
arrows  which  show  the  direc- 
tions are  marked  with  similar 
numerals  in  the  force  diagram. 
The  arrows  marked  1  in  Fig.  68  are  the  same  as  those  in  the  force 
diagram  of  Fig.  67. 

Since  the  force  bd  at  the  lower  joint  is  upward,  it  must  be 
downward  at  the  upper  joint.  An  arrow  pointing  downward  is 
now  placed  on  this  member  at  the  upper  joint,  as  shown  in  Fig. 
68.  In  the  line  bd  of  the  force  diagram  a  similar  arrow  is  placed 
and  marked  with  the  numeral  2.  A  line  is  then  drawn  through  b 
parallel  to  the  member  BC  of  the  space  diagram,  and  a  line  is 
drawn  through  d  parallel  to  DC  of  the  space  diagram.  The  inter- 


40  I  b. 

Space  Diagram 


Force  Diagram 


FIG.  68. 


CHAP.  IV]     CONCURRENT  CO-PLANAR  FORCES 


67 


section  of  these  lines  gives  the  point  c.  Since  the  arrow  marked  2 
in  the  force  diagram  is  from  b  toward  d,  the  arrow  in  dc must  point 
from  d  toward  c,  and  the  arrow  in  cb  must  point  from  c  toward  6. 
Finally,  the  corresponding  arrows  are  drawn  at  joint  2  of  the 
space  diagram. 

The  length  of  the  lines  on  the  final  force  diagram  are  now  meas- 
ured, and  the  results  expressed  in  pounds.  These  figures  are 
frequently  written  in  the  space  diagram.  All  the  stresses  in 
Fig.  68  are  tensile.  In  most  problems,  some  stresses  are  tensile 
and  some  are  compressive.  The  character  of  the  stress  is 
marked  on  the  space  diagram. 

It  is  understood  of  course,  that  Fig.  67  is  merely  one  step  of 
Fig.  68,  drawn  separately  for  clearness  of  explanation.  In 
solving  the  problems  below,  draw  a  single  space  diagram  and  a 
single  force  diagram,  as  in  Fig.  68.  The  numbered  arrows  are 
not  generally  used.  They  afford,  however,  great  help  to  begin- 
ners. It  is  best  to  draw  the  space  diagram  accurately  to  scale. 
The  directions  on  the  force  diagram  are  found  by  drawing  lines 
parallel  to  the  members  of  the  space  diagram. 

Problems 

1.  Solve  the  example  of  Fig.  67  graphically  to  the  scale  of  1  inch  =  10 
pounds,  marking  d  at  the  top  and  a  at  the  bottom  of  the  first  vertical  line. 
Compare  the  final  force  diagram  with  that  of  Fig.  68. 

2.  Solve  the  example  of  Fig.  66  graphically  to  the  scale  of  1  inch  =  1  foot 


340 


Force 

Diagram 


I  b. 


Space  D'ragram 


FIG.  69. 


in  the  space  diagram,  and  1  inch  =  40  pounds  in  the  force  diagram.  Scale 
the  force  diagram  and  put  the  results  in  pounds  on  the  space  diagram,  as 
in  Fig.  69. 

NOTE. — The  values  of  Fig.  69  were  taken  from  the  original  drawing  of 
the  force  diagram,  which  has  been  reduced  in  scale  for  printing. 


68 


MECHANICS 


[AKT.  53 


3.  The  mast  of  a  derrick  is  15  feet  long  and  the  boom  is  20  feet  long. 
One  guy  rope,  in  the  same  vertical  plane  as  the  boom,  makes  an  angle  of 
30  degrees  with  the  horizontal.  The  boom  makes  an  angle  of  15  degrees 
with  the  horizontal  and  carries  a  load  of  300  pounds.  Find  all  internal 
forces,  the  tension  in  the  guy  rope,  and  the  horizontal  and  vertical  compo- 
nents of  the  reaction  at  the  bottom  of  the  mast.  Use  the  scale  of  1  inch  =  5 
feet  on  the  space  diagram,  and  1  inch  =  100  pounds  on  the  force 
diagram. 


FIG.  70. 


NOTE. — In  an  actual  derrick,  the  boom  is  attached  to  the  mast  at  some 
distance  above  the  bottom.  The  cable  which  supports  the  load  runs  parallel 
to  the  boom,  and  increases  the  compression  in  that  member.  The  boom  is 
lifted  by  several  cables,  which  run  over  pulleys.  The  part  of  the  cable 
which  comes  down  the  mast  increases  the  compression  in  that  member. 

4.  Solve  Problem  3  by  moments  and  resolutions,  beginning  with  the 
right  end  of  the  boom  as  the  first  free  body.     Show  that  the  compression 
in  the  boom  is  the  same,  no  matter  what  angle  it  makes  with  the  horizontal. 

(In  the  solution  by  moments  and  resolutions,  it  is  convenient  to  represent 
each  force  by  a  single  letter,  as  H,  V,  P,  etc.,  instead  of  by  two  letters  as  in 
Bow's  method.) 

5.  Solve  Problem  3  graphically  when  the  boom  is  elevated  to  make  an 
angle  of  45  degrees  with  the  horizontal. 

53.  Summary. — The  resultant  of  concurrent,  coplanar  forces 
is  their  vector  sum. 

The  graphical  condition  of  equilibrium  for  a  set  of  concurrent, 
coplanar  forces,  is  that  the  force  diagram  is  a  closed  polygon. 

A  problem  of  the  equilibrium  of  concurrent,  coplanar  forces 
may  be  solved  by  any  one  of  the  following  methods : 

1 .  Construct  the  force  polygon  and  determine  the  magnitude  and 
direction  of  each  of  the  unknown  forces  by  measurement. 

2.  Construct  the  force  polygon  and  solve  trigonometrically.     This 
method  is  convenient  when  the  force  polygon  is  a  triangle,  espe- 
cially if  it  is  a  right-angled  or  an  isosceles  triangle. 

3.  Write  two  resolution  equations.     The  sum  of  the  components 
along  any  direction  is  zero. 


CHAP.  IV]     CONCURRENT  CO-PLANAR  FORCES  69 

4.  Write  two  moment  equations.     The  sum  of  the  moments  about 
any  point  is  zero.     The  two  origins   of  moment  and  the  point 
of  application  of  the  forces  must  not  lie  in  the  same  straight  line. 

5.  Write  one  moment  equation  and  one  resolution  equation.     The 
resolution  must  not  be  perpendicular  to  the  straight  line  which 
joins  the  origin  of  moments  and  the  point  of  application  of  the 
forces. 

Jointed  frames,  in  which  the  forces  are  parallel  to  the  members 
connecting  the  joints,  may  be  solved  as  a  series  of  problems  of 
concurrent  forces.  Begin  with  a  joint  at  which  there  are  only 
two  unknowns,  and  solve  by  any  of  the  methods  above.  By 
the  use  of  Bow's  method  of  lettering,  a  problem  of  this  kind  may 
be  solved  graphically,  with  all  the  forces  on  a  single  diagram. 

A  resolution  perpendicular  to  the  direction  of  an  unknown 
force  eliminates  that  force.  Moment  about  a  point  in  .the 
line  of  action  of  an  unknown  force  eliminates  that  force. 

The  moment  of  a  force  is  the  product  of  the  magnitude  of  the 
force  multiplied  by  the  component  of  the  apparent  arm  perpendicu- 
lar to  it;  or  the  product  of  the  entire  apparent  arm  multiplied 
by  the  component  of  the  force  perpendicular  to  it.  When  the 
coordinates  of  the  ends  of  the  apparent  arm  are  given,  instead  of 
its  length  and  direction,  it  is  advisable  to  resolve  both  the  force 
and  the  apparent  arm  into  their  components. 

M  =  Vx  -  Hy,  Formula  V. 

in  which  V  and  H  are  the  components  of  the  force  and  x  and  y 
are  the  components  of  the  arm.  The  component  V  is  parallel  to 
y  and  the  component  H  is  parallel  to  x. 

54.  Miscellaneous  Problems 

1.  Find  the  direction  and  magnitude  of  the  resultant  of  24  pounds  at  20 
degrees,  30  pounds  at  50  degrees,  20  pounds  at  110  degrees,  and  16  pounds 
at  210  degrees.     Solve  by  resolutions  along  one  pair  of  axes  at  right  angles 
to  each  other.     Check  by  resolutions  along  a  second  pair  of  axes. 

2.  A  mass  of  50  pounds  on  a  smooth  inclined  plane,  which  makes  an 
unknown  angle  with  the  horizontal,  is  held  by  a  force  of  18  pounds  parallel 
to  the  plane.     Find  the  inclination  of  the  plane  and  the  normal  reaction. 

y  3.  An  unknown  mass  is  placed  on  a  smooth  inclined  plane  which  makes 
an  unknown  angle  with  the  horizontal.  It  is  held  in  equilibrium  by  a  force 
of  18  pounds  parallel  to  the  plane,  or  by  a  force  of  20  pounds  at  an  angle 
of  10  degrees  above  the  horizontal.  Find  the  mass  of  the  body  and  the 
inclination  of  the  plane.  Also  find  the  normal  reaction  of  the  plane  in  each 
case, 


70  MECHANICS  [ART.  54 

4.  Solve  Problem  3  graphically. 

5.  A  40-pound  mass  slides  on  a  smooth  straight  rod  which  makes  an 
angle  of  25  degrees  with  the  horizontal.     It  is  attached  by  means  of  a 
weightless  cord  to  a  mass  of  20  pounds  which  slides  on  a  second  smooth 
rod.     This  rod  makes  an  angle  of  35  degrees  with  the  horizontal.     The  two 
rods  are  in  the  same  vertical  plane  and  on  opposite  sides  of  the  vertical. 
Find  the  direction  of  the  cord,  the  tension  which  it  exerts,  and  the  normal 
reaction  of  each  rod. 

Figure  71,  II,  shows  the  masses  supported  by  two  cords.  The  problem  is 
the  same  as  that  of  Fig.  71,  I,  and  the  diagram  is  more  convenient  for  a 
graphical  solution. 

Considering  the  40-pound  mass  as  the  free  body  in  equilibrium,  and 

resolving  horizontally, 

P  sin  25°  =  Q  cos  6.  (1) 

Considering  the  20-pound  mass  as  the 
free-body  in  equilibrium,  and  resolving 
horizontally, 

Q  sin  35°  =  T  cos  0.  (2) 


FIG.  72. 

Combining  Equations  (1)  and  (2) 

P  sin  25°  =  Q  sin  35°.  (3) 

Equation  (3)  is  the  same  as  if  the  forces  P  and  Q  were  concurrent. 
Resolving  vertically, 

P  cos  25°  =  40  -  T  sin  6,  (4) 

Q  cos  35°  =  20  +  T  sin  6,  (5) 

from  which 

P  cos  25°  +  Q  cos  35°  =  60.  (6) 

Equation  (6)  is  the  same  as  if  all  the  external  forces  were  concurrent. 

Combining  Equations  (3)  and  (6),  P  =  39.74  lb.;  Q  =  29.28  lb.     Com- 
bining Equations  (1)  and  (4),  6  =  13°  23',  T  =  17.23  lb. 

6.  Letter  Fig.  71,  II,  by  Bow's  method  and  solve  graphically. 

7.  A  mass  of  5  pounds  is  supported  by  a  cord  AB,  which  is  2  feet  long  and 
is  fixed  at  A,  and  by  a  second  cord  which  runs  over  a  smooth  pulley  and 
supports  a  mass  of  3  pounds.     The  point  (C  of  Fig.  72)  at  which  the  second 
cord  is  tangent  to  the  pulley  is  4  feet  from  A. in  a  horizontal  direction. 
Find  the  angle  which  AB  makes  with  the  horizontal  when  the  system  is 
in  equilibrium. 


CHAP.  IV]         CONCURRENT  CO-PLANAR  FORCES 


71 


Taking  moments  about  A  and  eliminating  trigonometrically,  the  equation 
is  found  to  be 

cos3  0  -  1.61  cos2  6  +  0.36  =  0 

This  equation  may  be  solved  by  the  trigonometric  method  for  the  solution 
of  a  cubic,  or  by  the  method  of  trial  and  error  (see  Art.  252).  One  root  is 
cos  6  =  0.5958.  To  get  the  remaining  two  roots  of  the  cubic,  divide  the 
equation  by  cos  6  —  0.5988,  and  solve  the  quadratic  thus  obtained. 

Two  of  these  roots  are  solutions  of  the  problem  of  mechanics.  (Construct 
the  space  diagram  for  one  of  these.)  The  third  root  is  a  solution  of  the 
mathematical  equation  but  is  impossible  mechanically. 

8.  A  20-pound  mass  is  supported  by  a  cord  AB  which  is  3  feet  long,  and 
a  second  cord  which  runs  over  a  smooth  pulley  2  feet  in  diameter  and  sup- 
ports a  mass  of  15  pounds.  The  axis  of  the  pulley  is  6  feet  to  the  right  and 


FIG.  73. 

1.5  feet  above  the  point  A,  the  upper  end  of  the  first  cord.  Find  the  posi- 
tion of  equilibrium. 

The  algebraic  solution  of  this  equation  is  difficult  and  involves  the  solu- 
tion of  an  equation  of  higher  degree  by  the  method  of  trial  and  error.  It  is 
most  easily  solved  by  a  combination  of  moments  and  graphics.  With  A 
as  the  center  and  with  a  radius  of  3  feet,  draw  the  arc  FG,  Fig.  73.  Select 
some  point  B  on  this  arc  and  draw  AB.  From  B  draw  a  line  tangent  to 
the  pulley.  The  length  AE  drawn  perpendicular  to  this  tangent  is  the 
moment  arm  of  the  15-pound  force.  The  moment  arm  of  the  20-pound 
force  is  AD.  For  equilibrium. 

20AD  =  15AE. 

If  the  moment  of  the  20-pound  force  is  too  great,  the  angle  6  is  too  small. 
Choose  another  point  on  the  arc  and  repeat  the  process.  Interpolate  for 
the  final  result. 


^  CHAPTER  V 

NON-CONCURRENT  CO -PLANAR  FORCES 

55.    Resultant  of  Parallel  Forces  in  the  Same  Direction. 

Figure  74  shows  a  common  case  of  parallel  forces  in  the  same  direc- 
tion. The  rigid  beam  ABC  carries  two  loads,  P  and  Q,  which 
are  parallel  forces.  The  equilibrant  of  these  two  forces  is  the 
force  S  at  B.  The  resultant  of  P  and  Q  is  a  force  at  B,  which  is 
equal  and  opposite  to  the  equilibrant.  The  direction,  magnitude, 


FIG.  75. 


and  position  of  the  resultant  may  be  found  by  first  finding  the 
equilibrant. 

To  find  the  equilibrant,  replace  the  rigid  beam  ABC  of  Fig. 
74  by  the  jointed  frame  of  Fig.  75.  Neglecting  the  weight  of  the 
frame,  the  forces  at  each  joint  may  be  found  by  the  methods  for 
concurrent  forces.  Figure  76, 1,  is  the  space  diagram  for  Fig.  75. 
For  simplicity,  each  member  is  represented  by  a  single  line.  Figure 

72 


CHAP.  V]    NON-CONCURRENT  CO-PLANAR  FORCES       73 

76,  II,  shows  the  force  diagram  constructed  by  Bow's  method. 
The  angle  a  is  assumed  to  be  known  and  the  angle  6  is  assumed  to 
be  unknown.  Beginning  at  joint  No.  1,  construct  the  force  tri- 
angle acd.  At  joint  No.  2,  the  direction  of  dc  is  from  left  toward 
right,  and  the  magnitude  of  cb  is  known.  The  force  triangle  is 
cbd.  The  direction  of  bd  gives  the  unknown  angle  0.  At 
joint  No.  3,  the  direction  and  magnitude  of  the  forces  ad  and 
db  are  known.  The  line  ba  represents  the  equilibrant  desired. 
The  force  ba  of  triangle  No.  3  is  along  the  same  line  as  the 
forces  P  and  Q,  and  is  equal  to  their  sum.  The  equilibrant  of 
two  parallel  forces  is  equal  to  the  sum  of  the  two  forces,  and  is  in 
the  opposite  direction. 

P  +  Q  +  S  =  0;P  +  Q=  -  S.  (1) 

The  resultant,  which  is  equal  and  opposite  the  equilibrant,  is 
the  algebraic  sum  of  P  and  Q  and  has  the  same  sign. 

P  +  Q  =  R  (2) 

The  force  polygon  of  Fig.  76,  II,  gives  the  direction  and  magni- 
tude of  the  equilibrant,  and,  consequently,  of  the  resultant. 
There  remains  the  problem  of  finding  algebraically  the  position 
of  the  equilibrant  or  resultant.  This  position  is  given  graphi- 
cally by  the  space  diagram  of  Fig.  76,  I.  It  is  now  desirable  to 
express  this  position  in  algebraic  language.  Extend  the  line  S 
of  Fig.  76,  I,  till  it  intersects  the  line  1-2  at  the  point  4.  Let 
the  distance  from  point  1  to  point  4  be  represented  by  x,  the  dis- 
tance from  point  2  to  point  4  be  represented  by  y,  and  the  dis- 
tance from  point  3  to  point  4  be  represented  by  z.  The  triangle 
1-3-4  of  the  space  diagram  is  similar  to  the  triangle  dac  of  the 
force  diagram;  and  the  triangle  2-4-3  of  the  space  diagram  is 
similar  to  the  triangle  deb  of  the  force  diagram.  From  the 
homologous  sides  of  1-3-4  and  dac, 

x  -  2-  m 

dc~P 

From  the  homologous  sides  of  2-4-3  and  deb, 


From  Equations  (3)  and  (4), 

X-Q-  (to 

y~P' 

Px  =  Qy.  Formula  VI. 

The  angles  a  and  6  do  not  appear  in  any  of  the  equations  above. 


74 


MECHANICS 


[ART.  55 


A-  cos<J> >t*--y  cos  6 


FIG.  77. 


It  is  evident,  therefore,  that  the  magnitude,  direction,  and  posi- 
tion of  the  equilibrant  (and  resultant)  are  independent  of  the 
form  of  the  jointed  frame.  These  are  the  same  whether  the 
rigid  body  is  a  jointed  frame,  as  in  Fig.  75,  or  a  single  beam,  as 
in  Fig.  74.  The  triangular  frame  is  merely  a  convenient  means 
of  finding  the  equilibrant. 

This  same  result  may  be  obtained  by  resolving  the  forces  P 
and  Q,  with  one  component  of  P  equal  and  opposite  to  one  com- 
ponent of  Q,  and  with  these  two  components  along  the  same  line. 

These  two  components  balance 
each  other.  The  remaining 
components  intersect.  Their 
resultant  at  -the  point  of  inter- 
section may  'be  found  by  the 
methods  for  concurrent  forces. 

In  Fig.  77,  a  line  is  drawn 
through  the  point  4  at  right 
angles  to  the  direction  of  P  and 
Q.  This  line  makes  an  angle  0  with  the  direction  of  the  line  1-2. 
Multiplying  both  sides  of  Formula  VI  by  cos  </>, 

Px  cos  0  =  Qy  cos  0.  (6) 

Since  x  cos  0  is  the  perpendicular  distance  from  the  point  4 
to  the  line  of  the  force  P,  the  term  P  x  cos  0  is  the  moment  of  the 
force  P  about  the  point  4  (or  about  any  point  in  the  line  of  the 
equilibrant).  Qy  cos  0  is,  likewise,  the  moment  of  the  force  Q 
about  point  4.  Equation  (6)  may  be  interpreted  as  follows: 
The  moment  of  one  of  the  parallel  forces  about  any  point  in  the  line 
of  the  resultant  is  equal  and  opposite  to  the  moment  of  the  other  force 
about  that  point. 

The  moment  of  the  result- 
ant of  two  parallel  forces 
about  any  point  in  their 
plane  is  equal  to  the  sum  of  ° "" 
the  moments  of  the  two  forces 
about  that  point.  This  may 
be  proved  from  Fig.  78. 

Take  moments  about  the  point  0  at  a  distance  e  from  the  line 
of  the  force  P  measured  in  the  direction  of  x  and  y.  The 
moment  of  the  resultant  about  this  point  is 

Mr  =  R(e  +  x)  cos  0  =  (Pe  +  Px  +  Qe  +  Qx)  cos  0.  (7) 


CHAP.  V]    NON-CONCURRENT  CO-PLANAR  FORCES       75 

The  sum  of  the  moments  of  P  and  Q  about  that  point  is  given  by 
the  equation, 

M  =  (Pe  +  Qe  -f  Qx  +  Qy)  cos  0.  (8) 

Subtracting  Equation  (8)  from  Equation  (7), 

Mr-  M  =  (Px  -  Qy)  cos  0  (9) 

Since  Px   =  Qy,  Mr  —  M  =  0,  and  Mr  =  M.     This  equation 
proves  the  proposition. 

If  there  are  three  forces  in  the  same  direction,  two  of  these 
forces  may  be  replaced  by  their  resultant.  This  resultant  is  then 
combined  with  the  third  force  to  get  the  resultant  of  all  three. 
This  process  may  be  continued  indefinitely.  The  resultant  of 
any  number  of  parallel  forces  in  the  same  direction  is  equal  to  the 
sum  of  the  forces,  and  the  moment  of  the  resultant  about  any  point 
is  equal  to  the  sum  of  the  moments  of  the  separate  forces  about  that 
point.  When  all  the  distances  are  measured  along  parallel  lines, 
the  term  cos  0  may  be  omitted  from  the  moment  equation. 

Example  I 

Two  forces  of  12  pounds  and  18  pounds  are  parallel,  in  the  same  direction, 
and  5  feet  apart.     Find  the  magnitude  and  position  of  their  resultant. 
Resultant  =  12  +  18  =  30  Ib. 

Taking  moments  about  a  point  in  the  line  of        |<- —-5'- ->| 

the  force  of  12  pounds,  Fig.  79,  [<- X »| ( 

12  X  0  =    0 
18  X  5  =  90 

30z  =  90  12  I  b 

x.=    3ft. 

The  resultant  is  a  force  of  30  pounds  at  a  dis-  R 

tance  of  3  feet  from  the  force  of  12  pounds.  FIG.  79. 


18  Ib. 


Example  II 

Given  the  following  forces,  all  of  which  are  vertically  downward:  8 
pounds  at  2  feet,  12  pounds  at  5  feet,  16  pounds  at  10  feet,  9  pounds  at  12 
feet,  and  5  pounds  at  14  feet.  Find  their  resultant. 

The  solution  is  best  arranged  in  this  form : 


Force,  Ib. 

Moment  arm,  ft. 

Moment,  ft.-lb. 

8 

2 

16 

12 

5 

60 

16 

10 

160 

9 

12 

108 

5 

14 

70 

50  414 

50z  =  4  14 

.   x  =  8.28  ft. 


76  MECHANICS  [AKT.  56 

The  resultant  is  50  pounds  at  a  distance  of  8.28  feet  from  the  line  of  the 
8-pound  force. 

Problems 

1.  Check  Example  II  by  moments  about  a  point  2  feet  from  the  8-pound 
force. 

2.  A  weightless,  horizontal  bar  is  12  feet  long  and  carries  40  pounds  at 
the  left  end,   55  pounds  3  feet  from  the  left  end,  80  pounds  6  feet  from 
the  left  end,  and  25  pounds  at  the  right  end.     Find  the  magnitude  of  the 
resultant  and  find  its  position  by  moments  about  the  left  end.     Check  by 
moments  about  some  other  point. 

3.  A  beam  20  feet  long  weighs  40  pounds.     Its  center  of  mass  is  9  feet 
from  the  left  end.     The  beam  is  placed  in  a  horizontal  position  and  loaded 
with  60  pounds  on  the  left  end,  50  pounds  6  feet  from  the  left  end,  80  pounds 
4  feet  from  the  right  end,  and  30  pounds  on  the  right  end.     The  beam  rests 
on  a  single  support.     Where  must  this  support  be  placed? 

4.  A  weightless  bar  5  feet  long  forms  a  part  of  a  jointed  frame  as  in  Fig. 
76.     The  bar  is  horizontal  and  carries  a  load  of  60  pounds  on  the  left  end 
and  a  load  of  40  pounds  on  the  right  end.     Solve  graphically  for  the  magni- 
tude and  position  of  the  equilibrant.     Make  the  angle  a  =  30°  and  use 
the  scale  of  1  inch  =  1  foot  on  the  space  diagram,  and  1  inch  =  20  pounds 
on  the  force  diagram. 

6.  Solve  Problem  4  with  a  =  45°. 

6.  Check  Problem  4  by  moments. 

7.  The  ends  of  a  bar  5  feet  in  length  are  connected  to  the  ends  of  a  chain 
which  is  7  feet  in  length.     A  load  of  50  pounds  is  hung  on  one  end  of  the 
bar  and  a  load  of  30  pounds  is  hung  on  the  other  end.     The  chain  is  suspended 
from  a  hook.     What  is  the  length  of  chain  between  the  hook  and  the  30 
pound  load  if  the  bar  hangs  horizontal? 

/  ;*V 
56.  Resultant    of    Parallel   Forces    in    Opposite  Directions. 

In  Fig.  76  or  Fig.  77,  the  force  Q  may  be  regarded  as  the  equilib- 
rant of  the  forces  P  and  S.     Numerically 


The  resultant  of  the  forces  P  and  S  in  opposite  directions  is 

the  force  RI  of  Fig.  80.     This  resultant  is  equal  and  opposite  the 

5  force  Q.     From  the  equations  of  the 

preceding  article,  it  is  evident  that  the 

__  I  _  jf    sum  of  the  moments  of  P,  Q,  and  S 
t  —  •       about  any  point  is  zero.     Since  RI  is 

equal  and  opposite  to  Q,  the  moment 

FlG.    80.  f    -n      -  i    .        xi  e    ,1 

of  RI  is  equal  to  the  sum  of  the  mo- 

ments of  S  and  P  taken  with  their  proper  signs.  The  moment 
of  the  resultant  of  two  parallel  forces  is  equal  to  the  sum  of  the 
moments  of  the  forces,  whether  the  forces  are  in  the  same  direction 
or  in  opposite  directions. 


CHAP.  V]     NON-CONCURRENT  CO-PLANAR  FORCES       77 

Example  I 

Find  the  magnitude  and  position  of  the  resultant  of  a  downward  force  of 
8  pounds  and  an  upward  force  of  18  pounds,  the  horizontal  distance  between 
the  lines  of  the  two  forces  being  5  feet. 

Taking  moments  about  a  point  in  the  line  of  the  8-pound  force,  and  giving 
the  negative  sign  to  the  downward  force, 

-8X0=    0 
18  X  5  =  90 
Wx  =  90 
x  =    9  ft. 

The  resultant  is  a  force  of  10  pounds  at  a  distance  of  9  feet  from  the  force 
of  8  pounds  and  4  feet  from  the  force  of  18  pounds.  The  resultant  is  up- 
ward. A  downward  force  of  10  pounds  along  the  same  line  would  balance 
the  upward  force  of  18  pounds  and  the  downward  force  of  8  pounds. 

The  resultant  of  any  number  of  parallel  forces  is  the  algebraic 
sum  of  the  forces;  the  moment  of  the  resultant  about  any  point  is 
the  algebraic  sum  of  the  moments  of  the  several  forces  about  that 
point. 

Example  II 

Find  the  resultant  of  10  pounds  down  at  2  feet,  17  pounds  down  at  5 
feet,  21  pounds  up  at  10  feet,  16  pounds  down  at  12  feet,  and  12  pounds 
down  at  15  feet.  Solve  by  moments  about  the  5-foot  position.  Call  down- 
ward direction  positive,  since  most  of  the  forces  are  downward.  With 
distances  toward  the  right  from  the  origin  regarded  as  positive,  clockwise 
moment  is  now  positive. 

Force,  Ib.  Arm,  ft.  Moment,  ft.-lb. 


10 

-3 

-30 

17 

0 

0 

21 

5 

-105 

16 

7 

112 

12 

10 

120 

34z 

97 

x  = 

2.85 

ft. 

Problems 

1.  Solve  Example  II  by  moments  about  the  0-foot  position. 

2.  Given  the  following  vertical  forces  in  the  same  vertical  plane:  20 
pounds  down  at  0  feet,  16  pounds  up  at  4  feet,  12  pounds  up  at  8  feet,  and 
10  pounds  down  at  10  feet.     Find  the  magnitude  and  position  of  the  re- 
sultant and  check. 

3.  A  horizontal  beam  20  feet  in  length  weighs  240  pounds.     Its  center 
of  mass  is  at  the  middle  of  its  length.     It  carries  a  load  of  160  pounds  5  feet 
from  the  left  end,  and  a  load  of  120  pounds  at  the  right  end.     The  left  end 


78  MECHANICS  [ART.  57 

rests  on  a  platform  scale  which  reads  170  pounds.     There  is  a  second  sup- 
port near  the  right  end.     Where  is  it  located,  and  what  load  does  it  carry? 
4.  By  the  graphical  method,  find  the  resultant  of  a  downward  force  of 
6  pounds  and  an  upward  force  of  10  pounds  at  a  distance  of  4  inches  apart. 

57.  Equilibrium  of  Parallel  Co-planar  Forces. — Since  the 
moment  of  the  resultant  of  several  parallel  forces  is  equal  to  the 
sum  of  the  moments  of  the  separate  forces,  the  moment  of  the 
equilibrant  is  equal  and  opposite  to  this  sum.  Consequently,  if 
the  body  is  in  equilibrium,  the  algebraic  sum  of  the  moments  of 
all  the  forces  is  zero,  and  the  sum  of  the  forces  is  zero.  These 
relations  may  be  expressed  by  the  equations, 

ZP  =  0; 

ZM  =  0. 

There  are  two  unknowns.  The  problem  may  be  solved  by  one 
moment  equation  and  one  resolution  parallel  to  the  direction  of 
the  forces.  The  resolution  equation  may  be  replaced  by  a  second 
moment  equation. 

Example  „ 

A  horizontal  beam  is  20  feet  long.  It  weighs  240  pounds  and  its  center 
of  mass  is  at  the  middle  of  its  length.  The  beam  is  supported  at  the  left 
end  and  at  4  feet  from  the  right  end.  It  carries  a  load  of  120  pounds  6  feet 
from  the  left  end,  and  a  load  of  160  pounds  at  the  right  end.  Find  the  reac- 
tion at  each  support. 


Ib 


FIG.  81. 

Take  moments  about  the  left  support,  in  order  to  eliminate  the  reaction 
Ri,  Fig.  81.  Call  downward  force  positive  and  distance  toward  the  right 
positive. 

Force,  Ib.                     Arm,  ft.  Moment,  ft.-lb. 

240                                10  2400 

120           0  720 

160          20  3200 


<-"-~<5-~-->r~ 

ieo  jib. 

-t—'- 

I       160. 

W^I^^=-    j^Es     * 

2 

j 

r 

»  X 

520  6320 

=  6320, 
=  395  Ib. 


CHAP.  V]    NON-CONCURRENT  CO-PLANAR  FORCES        79 

Take  moments  about  the  right  support  and  call  distance  toward  the  left 
positive. 

Force,  Ib.  Arm,  ft.  Moment,  ft.-lb. 

240  6  1440 

120  10  1200 

160  -4  -640 


520  2000 

16fli  =  2000, 

R!  =     125  Ib. 
Check  by  vertical  resolutions;  395  +  125  =  520  Ib. 

Problems 

1.  A  horizontal  beam  24  feet  long,  with  its  center  of  mass  at  the  middle, 
weighs  180  pounds.     It  is  supported  4  feet  from  the  left  end  and  2  feet  from 
the  right  end.     It  carries  144  pounds  on  the  left  end,  160  pounds  13  feet 
from  the  left  end,  162  pounds  6  feet  from  the  right  end,  and  90  pounds  on 
the  right  end.     Find  the  reactions  at  the  supports  and  check. 

2.  The  beam  of  Problem  1  is  supported  2  feet  from  the  left  end.     A  second 
support  near  the  right  end  exerts  an  upward  force  of  400  pounds.     Where  is 
this  second  support  located,  and  what  is  the  reaction  of  the  left  support? 

3.  A  beam  20  feet  long,  with  its  center  of  mass  at  the  middle  of  its  length, 
weighs  160  pounds.     It  is  supported  5  feet  from  the  left  end  and  is  held 
down  by  a  force  one  foot  from  the  left  end.     The  beam  carries  a  load  of 
120  pounds  on  the  right  end.     Find  the  reactions  and  check. 

58.  Condition  of  Stable  Equilibrium. — When  a  body  is  in 
equilibrium  under  the  action  of  non-concurrent  forces,  insofar  as 
the  equilibrium  in  that  position  is  concerned,  any  one  of  the 
forces  may  be  regarded  as  applied  to  the  body  at  any  point  along 
its  line  of  action.  If,  however,  the  equilibrium  be  disturbed, 
either  by  a  slight  displacement  of  the  body,  or  by  a  small  change 
in  the  magnitude  of  one  of  the  forces,  then  the  position  of  applica- 
tion of  the  forces  is  the  ele- 
ment which  determines 
whether  the  body  will  take  a 
new  position  with  a  small  dis- 
placement, or  will  continue  to 
move  through  a  greater  dis- 
tance. The  position  of  ap- 
plication of  the  forces  determines  whether  the  equilibrium  is 
stable,  unstable,  or  neutral. 

Figure  82  shows  a  rigid  body  which  is  supported  at  B  and  loaded 
at  A.  A  second  load.Q  acts  along  the  line  through  C%,  C  and  C\. 
If  the  body  is  in  equilibrium  in  this  position,  it  is  in  equilibrium 


80 


MECHANICS 


[ART.  58 


whether  the  load  is  applied  at  C,  at  Ci,  or  at  C2.  The  points 
A,  B,  and  C  lie  in  a  straight  line,  as  shown  in  Fig.  83, 1.  If  the 
beam  is  turned  through  an  angle  0  from  the  horizontal  position, 
the  moment  arm  of  the  force  P  becomes  x  cos  B  and  the  moment 
arm  of  Q  becomes  y  cos  B.  In  the  original  position  of  equilibrium 


consequently 


Px  =  Qy; 
Px  cos  B  =  Q  y  cos  B. 


There  is  the  same  relative  change 
in. the  moments  of  the  two  forces; 
therefore,  the  beam  is  in  equi- 
librium in  the  new  position. 

When  a  body  is  in  equilibrium 
under  the  action  of  three  parallel 
forces,  if  the  points  of  application 
of  the  three  forces  lie  in  a  straight 
line,  the  equilibrium  is  neutral. 

Figure  83,  II,  represents  the 
case  in  which  the  load  is  applied 
at  a  point  Ci  placed  above  the 
line  through  A  and  B.  The 
broken  lines  illustrate  the  con- 
dition when  the  beam  is  rotated 
in  a  clockwise  direction  about  B. 
The  moment  arm  of  Q  becomes 

longer  and  the  moment  arm  of  P  becomes  shorter.  After  the  arm 
BC  has  passed  the  horizontal  position,  the  moment  arm  of  Q 
diminishes,  but  less  rapidly  than  the  effective  arm  of  the  force 
P.  If  the  forces  P  and  Q  in  the  original  position  produced  equal 
and  opposite  moments  about  B,  the  moment  of  Q  in  any  one  of 
these  new  positions  is  greater  than  the  moment  of  P,  and  the 
beam  continues  to  rotate  about  B  through  approximately  180 
degrees  to  the  position  of  stable  equilibrium.  The  equilibrium 
of  Fig.  83,  II,  is  unstable.  If  the  beam  had  been  rotated  in  the 
opposite  direction  from  the  position  of  equilibrium,  the  same 
condition  would  obtain  and  it  would  continue  to  rotate  in  that 
direction. 

Figure  83,  III,  represents  the  case  in  which  the  load  is  applied  at 
C2  placed  below  the  line  through  A  and  B.  When  the  beam  is 
rotated  slightly  from  this  position  in  a  clockwise  direction,  the 


FIG.  83. 


" 


r\c 


CHAP.  V]     NON-CONCURRENT  CO-PLANAR  FORCES       81 

moment  arm  of  P  becomes  a  little  longer  and  that  of  Q  becomes 
a  little  shorter.  The  beam  will  return  to  its  original  position 
after  displacement.  The  equilibrium  is  stable. 

Example 

Figure  84  represents  a  rectangular  board  in  a  vertical  plane  supported  by  a 
smooth  hinge  at  the  middle.  A  load  of  10  pounds  is  suspended  from  A2 
and  a  load  of  Q  pounds  is  sus- 
pended from  C2.  The  board  ro- 
tates through  an  angle  of  10 
degrees  in  a  clockwise  direction. 
Find  the  load  Q. 

It  is  assumed  that  the  center  of 
mass  is  at  the  center  of  the  board, 
so  that  its  weight  exerts  no  mo-  ^IG-  &*• 

ment  about  the  support  in  any  position.  Taking  moments  about  the 
support, 

10  X  13  cos  12°  37'  =  Q  X  13  cos  32°  37';  or 
10  (12  cos  10°  +  5  sin  10°)   =  Q  (12  cos  10°  -  5sin  10°) 

•  % 
Problems 

1.  In  Fig.  84,  the  load  at  A  2  is  10  pounds  and  the  load  at  C2  is  12  pounds. 
Find  the  position  of  equilibrium.     Ans.  The  rotation  is  12°  36'  clockwise. 

2.  Solve  Problem  1  if  the  loads  are  20  pounds  and  22  pounds,  respectively. 

3.  In  Fig.  84,  the  board  weighs  4  pounds  and  its  center  of  mass  is  2  inches 
below  the  point  of  support.     The  load  P  is  attached  at  A  and  the  load  Q  at 
C.     The  points  A,  B,  and  C  are  in  a  horizontal  straight  line.     Find  the 
position  of  equilibrium  if  P  is  10  pounds  and  Q  is  11  pounds  and  there  is  no 
friction  at  B.  Ans.     Rotation  =  tan-1  -f  =  56°  19'. 

4.  Solve  Problem  3  for  loads  of  5  pounds  and  6  pounds,  respectively. 

6.  In  Fig.  84,  the  loads  are  suspended  from  AI  and  Ci,  which  are  1  inch 
above  A  and  C,  respectively.  Find  the  position  of  equilibrium  when  P  is 
1  pound  and  Q  is  2  pounds.  Ans.  Rotation  =  tan-1  2.4  =  67°  23'. 

6.  Solve  Problem  5  for  loads  of  4  pounds  and  5  pounds. 

7.  In  Fig.  84,  a  load  of  10  pounds  is  applied  at  A  2  and  a  load  of  11  pounds 
at  C2.     The  board  weighs  4  pounds  and  its  center  of  mass  is  2  inches  below 
B.     Find  the  position  of  equilibrium.  Ans.     Rotation  =  6°  04'. 

8.  Solve  Problem  7  if  the  loads  are  1  pound  and  2  pounds. 

59.  Resultant  of  Non -parallel  Forces  Graphically. — Figure  85 
shows  three  forces,  which  are  supposed  to  be  applied  to  a  rigid 
body.  These  forces  are:  20  pounds  at  30  degrees  with  the 
horizontal,  applied  at  the  point  (5,  0) ;  15  pounds  at  75  degrees 
with  the  horizontal,  applied  at  the  point  (3,  2) ;  and  12  pounds  at 
115  degrees  with  the  horizontal  toward  the  right,  applied  at  the 


82 


MECHANICS 


[ART.  GO 


point  (  —  2,3).  It  is  desired  to  find  the  magnitude  and  direction 
of  the  resultant,  and  the  position  of  the  line  along  which  it  acts. 
Figure  85,  II,  is  the  force  diagram.  The  forces  of  20  pounds  and 
15  pounds  are  laid  off  and  their  resultant  is  found.  This  resultant 
isRi.  Then  on  the  space  diagram  the  lines  of  the  20-pound  force 
and  the  15-pound  force  are  extended  till  they  intersect  at  B. 
Through  B  a  line  is  drawn  parallel  to  the  direction  of  Ri  of  the 


\ 

3 

/ 

,2) 

\ 

2 

1 
1 

V     1 
\ 

I 

2 

1  3 

•4 

5 

x 

x 

-3 

-a 

-' 

\ 
\ 

R/' 

! 

> 

-X 

(5 

o) 

-' 

'p 

-2 

\ 

// 

\ 

-3 

I 

C 

FIG.  85. 

force  diagram.  The  resultant  of  the  forces  of  20  pounds  and  15 
pounds  may  be  regarded  as  acting  along  this  line.  A  vector 
representing  the  force  of  12  pounds  is  now  added  to  Ri  of  the 
force  diagram.  The  vector  sum  is  R,  which  represents  the  direc- 
tion and  magnitude  of  the  resultant  of  all  three  forces.  On  the 
space  diagram,  the  line  of  the  12-pound  force  is  extended  till  it 
intersects  the  line  of  RI  at  C.  Through  C  a  line  is  drawn  parallel 
to  R  of  the  force  polygon.  This  line  gives  the  position  of  the 
resultant  of  all  three  forces. 

Problem 

Given  the  following  forces :  20  pounds  horizontal  toward  the  right  through 
the  point  (2,  4);  25  pounds  at  an  angle  of  35  degrees  to  the  right  of  the 
vertical  through  the  point  (3,  3);  15  pounds  vertical  through  the  point  (1,  3); 
and  16  pounds  at  an  angle  of  45  degrees  to  the  left  of  the  vertical  through 
the  point  (—2,  2).  Find  the  direction  and  magnitude  of  the  resultant  by 
means  of  the  force  polygon,  and  find  its  position  on  the  space  diagram. 
Use  1  inch  =  5  pounds  on  the  force  diagram,  and  1  inch  =  2  units  of  length 
on  the  space  diagram.  Measure  the  perpendicular  distance  from  the  point 
(0,  0)  to  the  line  of  the  resultant. 

60.  Calculation  of  the  Resultant  of  Non-parallel  Forces. — The 
force  polygon  of  Fig.  85,  II,  is  exactly  the  same  as  that  for  a  set 
of  concurrent  forces.  It  is  evident,  therefore,  that  the  magnitude 
and  direction  of  the  resultant  of  non-concurrent  forces  may  be 
calculated  by  the  method  of  Art.  42.  Each  force  is  resolved  into 


CHAP.  V]    NON-CONCURRENT  CO-PLANAR  FORCES       83 

two  components  along  a  pair  of  axes  at  right  angles  to  each  other. 
The  sum  of  the  components  along  one  axis  is  taken  as  one  side  of 
right-angled  triangle.  The  sum  of  the  components  along  the  other 
axis  is  taken  as  the  other  side  of  the  same  right-angled  triangle. 
The  resultant  is  represented  by  the  hypotenuse  of  his  triangle. 

The  position  of  the  line  of  action  of  this  resultant  is  calculated 
by  moments.  In  Fig.  85,  the  force  of  20  pounds  and  the  force  of 
15  pounds  may  be  regarded  as  concurrent  at  B.  According  to 
Art.  48,  the  moment  of  RI  about  any  point  is  equal  to  the  sum  of 
the  moments  of  the  20-pound  force  and  the  15-pound  force  about 
that  point.  The  resultant  RI  may  be  regarded  as  concurrent 
with  the  force  of  12  pounds  at  C.  The  sum  of  the  moments  of 
RI  and  the  12-pound  force  about  any  point  is  equal  to  the  mo- 
ment of  the  final  resultant  about  that  point.  The  moment  about 
any  point  of  the  resultant  of  a  set  of  non-concurrent,  coplanar 
forces  is  equal  to  the  sum  of  the  moments  of  the  several  forces  about 
that  point. 

Example 

A  rectangular  board,  Fig.  86,  is  4  feet  wide  and  3  feet  high.  A  force  of 
20  pounds,  at  an  angle  of  15  degrees  with  the  horizontal  toward  the  right, 


ni 


FIG.  86. 

is  applied  at  the  lower  right  corner.  A  force  of  24  pounds,  at  an  angle  of 
30  degrees  to  the  right  of  the  vertical,  is  applied  at  the  upper  right  corner. 
A  force  of  16  pounds  at  an  angle  of  45  degrees  to  the  left  of  the  vertical  is 


84  MECHANICS  [ART.  60 

applied  at  the  upper  left  corner.     Find  the  magnitude  and  direction  of  the 
resultant,  and  its  distance  from  the  lower  left  corner. 
Resolving  horizontally  and  vertically, 

Force  H  component  V  component 

20  19.318  5.176 
24  12.000  20.784 
16  -11.314  11.314 


20.004         37.274 

The  resultant  is  the  hypotenuse  of  the  right-angled  triangle  of  which  the 
base  is  20.004  units  and  the  altitude  is  37.274  units.  The  resultant  makes 
an  angle  of  61°  47'  with  the  horizontal  toward  the  right.  Its  magnitude  is 
42.30  pounds. 

To  find  the  location  of  the  line  of  the  resultant  force  on  the  space  diagram, 
moments  are  taken  about  the  lower  left  corner  of  the  board.  Since  the 
horizontal  and  vertical  components  of  the  several  forces  have  already  been 
computed,  it  is  convenient  to  use  these  components  in  calculating  the 
moments. 

The  horizontal  component  of  -the  20-pound  force  and  the  vertical  com- 
ponent of  the  16-pound  force  pass  through  the  origin  of  moments  so  that 
the  moment  of  each  of  these  components  is  zero. 

V  component  of  20  Ib.  5 . 176  X  4  =       20 . 704 

V  component  of  24  Ib.  20 . 784  X  4  =       83 . 136 

#  component  of  24  Ib.  12.000  X  3  =  -  36.000 

H  component  of  16  Ib.  11 . 314  X  3  =       33. 942 


M  =     101 .  782  ft.-lb. 


Figure  86,  III,  shows  the  location  of  the  resultant.  Since  the  resultant 
makes  an  angle  of  61°  47'  with  the  horizontal,  a  line  perpendicular  to  it  makes 
an  angle  of  28°  13'  with  the  horizontal.  To  locate  the  resultant  on  the  space 
diagram,  a  line  is  first  drawn  through  the  lower  left  corner  at  an  angle  of 
28°  13'  below  the  horizontal  toward  the  right.  A  length  of  2.406  feet  is 
measured  off  on  this  line  from  the  origin  of  moments  to  the  point  B.  The 
moment  is  counter-clockwise  and  the  resultant  force  is  upward;  hence  the 
moment  arm  must  be  measured  toward  the  right  from  the  corner  of  the 
board.  Through  the  point  B,  a  line  is  drawn  perpendicular  to  OB.  The 
resultant  force  is  applied  along  this  perpendicular  line.  If  a  smooth  pin 
were  passed  through  the  board  at  any  point  along  this  line,  this  pin  would 
hold  the  board  in  equilibrium,  and  the  reaction  at  the  point  of  contact 
would  be  42.30  pounds.  The  equilibrium  would  be  stable,  unstable  or 
neutral,  depending  upon  the  position  of  the  pin  in  the  line  of  action  of  the 
resultant  force. 


CHAP.  V]    NON-CONCURRENT  CO-PLANAR  FORCES       85 

It  is  best  to  determine  the  sign  of  the  moments  by  observing  on  the  space 
diagram  whether  the  rotation  is  clockwise  or  counter-clockwise,  rather  than 
by  using  the  signs  of  the  forces  and  effective  arms.  A  horizontal  force 
toward  the  right  applied  to  a  vertical  arm  upward  gives  a  clockwise  moment. 
A  vertical  force  upward  applied  to  a  horizontal  arm  toward  the  right  gives 
a  counter-clockwise  moment.  If  counter-clockwise  moment  be  taken  as 
positive, 

M  =  Vx  -  Hy, 

in  which  V  is  the  component  of  the  force  parallel  to  the  Y  axis,  H  is  the 
component  of  the  force  parallel  to  the  X  axis,  x  is  the  abscissa,  and  y  is  the 
ordinate  of  some  point  in  the  line  of  action  of  the  force.  It  is  not  necessary 
that  the  X  axis  be  horizontal  and  the  Y  axis  vertical.  The  X  axis  may  have 
any  direction  and  the  F  axis  be  perpendicular  to  the  X  axis. 

The  student  who  prefers  to  make  use  of  the  signs  of  the  forces  and  dis- 
tances may  employ  the  equation  above,  but  he  should  also  check  his  results 
and  visualize  his  problem  by  observing  the  direction  of  rotation  on  the  space 
diagram. 

Instead  of  finding  the  perpendicular  distance  from  the  origin  of  moments 
to  the  line  of  the  resultant,  the  points  of  intersection  of  the  resultant  force 
with  the  axes  of  coordinates  might  have  been  calculated.  Since  the  origin 
of  moments  lies  in  the  X  axis,  the  moment  of  the  horizontal  component  is 
zero  at  the  point  where  the  line  of  the  resultant  cuts  that  axis. 

101.782  =  Vx, 
in  which  x  is  the  abscissa  of  the  point  of  intersection  with  the  X  axis. 

_  101.782  _ 
~37^74~ 

Regarding  the  force  as  applied  at  the  Y  intercept, 

101.782 


The  moment  is  counter-clockwise.  A  vertical  force  upward  gives  a  counter- 
clockwise moment  when  the  point  of  application  is  to  the  right  of  the  origin 
of  moments.  A  horizontal  force  toward  the  right  gives  a  counter-clockwise 
moment  when  the  point  of  application  is  below  the  origin. 

Problems 

1.  Solve  the  Example  of  Fig.  85  by  resolutions  and  moments. 

2.  Solve    the   Problem   of   Art.    59  and   compare  with  your  graphical 
solution. 

3.  Find  the  direction,  magnitude,  and  the  line  of  application  of  the  resul- 
tant of  the  following  forces  :  20  pounds,  at  65  degrees  to  the  right  of  the  ver- 
tical, through  the  point  (2,  3);  16  pounds,  at  20  degrees  to  the  right  of  the 
vertical,  through  the  point  (—2,  2);  12  pounds,  at  40  degrees  to  the  left  of 
the  vertical,  through  the  point  (1,  5);  and  18  pounds,  at  110  degrees  to  the 
left  of  the  vertical,  through  the  point  (—3,  2). 


86 


MECHANICS 


[ART.  61 


The  solution  may  be  arranged  in  tabular  form  as  was  done  with  Problem 
4  of  Art.  17. 


Force, 
Ib. 

Angle, 
deg. 

X 

y 

cos  ot 

sin  a 

H 

V 

Vx 
(counter-cl 

-Hy 
ockwise  +) 

20 

25 

2 

3 

0.9063 

0.4226 

18.126 

8  452 

16.904       -54.378 

16 

70 

-2 

2 

0.3420 

0.9397 

5.472 

15.035 

-30.070 

-10.944 

12 

130 

1 

5 

-0.6428 

0.7660 

-    7.714 

9.192 

9.192 

38.570 

18 

200 

-3 

2 

-0.9397 

-0.3420 

-16.915 

-6.156 

18.468 

33.830 

-    1.031 

26.523 

14.494 

7.078 

14.494 

Total  moi 

nent  

21.572 

!             1 

The  resultant  force  is  26.55  pounds  at  an  angle  of  2°  14'  to  the  left  of  the 
vertical.  The  total  moment  is  21.572  units  counter-clockwise.  The  resul- 
tant force  passes  through  a  point  at  a  distance  of  0.812  units  from  the  origin, 
measured  along  a  line  which  makes  an  angle  of  2°  14'  above  the  horizontal 
toward  the  right. 

4.  In  Problem  3,  find  the  point  where  the  resultant  cuts  the  X  axis,  by 
dividing  the  total  moment  by  the  vertical  component  of  the  resultant  force. 

Ans.  x  =  0.813. 

5.  In  Problem  3,  find  the  point  where  the  resultant  cuts  the  Y  axis. 

6.  In  Problem  3,  compute  the  moments  about  the  point  (2,  2),  instead 
of  the  origin,  and  find  the  points  of  intersection  of  the  resultant  force  with 
the  lines  x  —  2  and  y  =  2. 


61.  Equilibrium  of  Non-concurrent  Forces. — A  set  or  non- 
concurrent  forces  acting  on  a  rigid  body  may  be  divided  into  two 
groups.  In  order  that  equilibrium  may  exist,  two  conditions 
must  be  satisfied.  These  are: 

1.  The  resultant  of  the  forces  of  one  group  must  be  equal  and 
opposite  to  the-  resultant  of  all  the  other  forces. 

2.  The  resultant  of  the  forces  of  one  group  must  lie  along  the 
same  line  as  the  resultant  of  all  the  other  forces. 

The  first  condition  is  identical  with  the  condition  of  equilibrium 
of  concurrent  forces.  Stated  graphically,  this  means  that  the 
force  polygon  must  close.  Stated  algebraically,  this  means  that 
the  sum  of  the  components  of  all  the  forces  along  any  direction 
is  zero.  Literally  it  is  written, 

PI  cos  on  +  P2  cos  az  +  PS  cos  «3  +  etc.  =  0,  (1) 

SP  cos  a  =  0.  (2) 


CHAP.  V[    NON-CONCURRENT  CO-PLANAR  FORCES       87 

In  the  case  of  coplanar  forces,  two  independent  equations  of  this 
kind  may  be  written. 

The  resultant  of  one  group  of  forces  may  be  equal  and  opposite 
to  the  resultant  of  all  the  other  forces,  but  may  not  lie  in  the 
same  line.  In  the  latter  case,  the  moment  of  one  group  of  forces 
will  not  balance  the  moment  of  the  other  forces,  and  the  forces 
will  rotate  the  body.  The  second  condition  is  satisfied  when 
the  sum  of  the  moments  of  all  the  forces  which  act  on  the  rigid 
body  is  zero.  This  condition  is  expressed  algebraically  by  the 
equation 

=  0.  (3) 


A  problem  of  the  equilibrium  of  non-concurrent,  coplanar 
forces  may  be  solved  by  writing  two  resolution  equations  and  one 
moment  equation.  Since  a  moment  equation  may  replace  a  resolu- 
tion equation,  any  one  of  the  following  combinations  may  be 
used  : 

1.  One  moment  equation  and  two  resolution  equations. 

2.  Two  moment  equations  and  one  resolution  equation. 

3.  Three  moment  equations. 

As  in  the  case  of  concurrent  forces,  resolution  perpendicular  to 
the  line  of  action  of  a  force  eliminates  that  force;  moment  about 
a  point  in  the  line  of  action  of  a  force  eliminates  that  force. 
Moment  about  the  point  of  intersection  of  two  forces  eliminates 
both  of  them.  When  two  forces  of  unknown  magnitude  inter- 
sect, it  is  advisable  to  begin  the  solution  by  writing  the  moment 
equation  about  their  point  of  intersection  as  the  origin.  Fre- 
quently, both  the  direction  and  magnitude  of  one  force  are  un- 
known. If,  however,  some  point  on  the  line  of  action  of  this 
force  is  known,  it  is  advisable  to  take  moments  about  this  point. 

Example 

A  bar  A  B,  Fig.  87,  is  20  feet  long,  weighs  60  pounds,  and  has  its  center 
of  mass  8  feet  from  A.  The  bar  is  hinged  at  A  and  supported  by  a  cord  at 
B.  The  bar  makes  an  angle  of  15  degrees  above  the  horizontal  toward  the 
right,  and  the  cord  makes  an  angle  of  35  degrees  to  the  left  of  the  vertical. 
Find  the  tension  in  the  cord  and  the  direction  and  magnitude  of  the  hinge 
reaction  at.il. 

The  free  body  is  the  entire  bar  AB.  The  forces  which  act  on  the  free 
body  are  its  weight,  the  unknown  tension  in  the  cord  at  B,  and  the  reaction 
of  the  hinge  at  A.  The  hinge  reaction  is  unknown,  as  to  both  direction 
and  magnitude. 


88 


MECHANICS 


[ART.  61 


Begin  by  writing  a  moment  equation  about  the  hinge  A,  since  this  elimi- 
nates two  unknowns. 


60  X  8  X  cos  15°  =  P  X  20  cos  20° 
24  X  0.9659  =  0.  9397P 
P  =  24.671b. 


(4) 


To  find  the  direction  and  magnitude  of  the  reaction  at  the  hinge,  it  is 
convenient  to  regard  it  as  made  up  of  a  horizontal  component  H  and  a 
vertical  component  V.  Resolving  horizontally, 


H  =  P  sin  35°  =  14.15  Ib. 
Resolving  vertically, 

V  =  60  -  P  cos  35°  =  60  -  20.21  = 


39.79  Ib. 


(5) 


(6) 


From  the  force  triangle,  Fig.  87,  II,  the  resultant  reaction  of  the  hinge  is 
found  to  be  42.23  pounds  at  an  angle  of  19°  35'  to  the  right  of  the  vertical. 
Check  all  results  by  moments  about  B, 


60  X  12  cos  15°  =  42.23  X  20  cos  34°  35' 


(7) 


Problems 

1.  Complete  the  check  of  the  Example  above  by  substituting  in  Equa- 
tion (7). 

2.  Solve  the  Example  of  Fig.  87  if  the  cord  makes  an  angle  of  35  degrees 
to  the  right  of  the  vertical.     Check  the  result. 

3.  Consider  the  hinge  reaction  in  Fig.  87  as  made  up  of  a  component 
parallel  to  AB  and  a  component  perpendicular  to  AB.     Find  the  component 
perpendicular  to  A  B  by  moments  about  B.     Find  the  component  parallel 
to  AB  by  a  resolution  parallel  to  the  bar.     Calculate  the  resultant  reaction 
and  check. 

4.  A  vertical  door,  Fig.  88,  is  12  feet  wide,  10  feet  high,  and  weighs  360 
pounds.     The  hinges  are  one  foot  from  the  top  and  bottom,  respectively, 
and  are  so  placed  that  all  the  vertical  load  comes  on  the  lower  hinge.     Find 
the  direction  and  magnitude  of  each  hinge  reaction. 

6.  A  derrick  mast  is  30  feet  long.     The  boom  is  &0  feet  long,  is  elevated 


.. ,-.  r 


CHAP.  V]    NON-CONCURRENT  CO-PLANAR  FORCES       89 

20  degrees  above  the  horizontal,  and  carries  a  load  of  1200  pounds.  A  guy 
rope  attached  to  the  top  of  the  mast  makes  an  angle  of  35  degrees  with  the 
horizontal.  Considering  the  mast,  boom,  and  the  ropes  which  connect  them 
as  a  single  rigid  body,  find  the  tension  in  the  guy  rope,  and  the  direction 
and  magnitude  of  the  reaction  at  the  base  due  to  the  load  of  1200  pounds. 

6.  A  derrick  mast  is  15  feet  long. 
The  boom  is  20  feet  long  and  is  ele- 
vated  to  a  position  in  which  the  free 
end   is   16  feet  from  the  top  of  the 
mast.     A  guy  rope  in  the  same  plane 
as  the  boom  and  mast  is  attached  to 
a  point  at  the  same  level  as  the  bot- 
tom of  the  mast,  at  a  distance  of  18 
feet  from  the  bottom.     Find  the  ten- 
sion  in   the  guy  rope  and  the  hori- 
zontal and  vertical  components  of  the 
reaction   at  the  bottom  of  the  mast 
due  to  a  load  of  1600  pounds  on  the 

end  of  the  boom.  Construct  the  space  diagram  to  the  scale  of  1  inch  =  5 
feet,  and  solve  by  measuring  moment  arms  on  the  diagram. 

7.  In  Problem  6,  calculate  the  angles  and  the  moment  arms  trigonomet- 
rically,  and  solve  by  one  moment  and  two  resolution  equations. 

8.  Solve  Problem  6  for  all  forces  by  Bow's  method,  drawing  the  lines  in 
the  force  diagram  parallel  to  the  corresponding  members  of  the  space 
diagram. 

9.  Solve  Problem  5  for  all  forces  joint  at  a  time  by  resolutions. 

10.  A  bar  A  B,  Fig.  89,  l  is  7  feet  long,  weighs  28  pounds,  and  has  its  center 
of  mass  3  feet  from  A.     The  end  A  is  provided  with  a  cylindrical  roller, 


»/. 

•I'1'1 

p 

? 

rv-S-T 

^ 

JSL. 

_»J 

—  T- 

..;, 

s' 

v\ 

't. 

3 

60 

# 

10' 

i  

Ji^ 

»|- 

•7 

i 

FIG.  88. 


which  allows  it  to  move  on  a  surface  with  little  friction.     The  bar  is  placed 
with  the  end  A  on  a  horizontal  platform.     It  is  supported  by  a  cord  BD 

1  The  apparatus  as  shown  in  Fig.  89  is  decidedly  unstable.  One  of  the 
masses  P  or  Q  should  be  free  to  move  only  a  short  distance.  The  mass 
might  be  placed  on  a  second  platform  scale  and  the  difference  of  weight 
taken.  To  avoid  lateral  instability,  the  pulley  at  D  should  be  to  the  right 
of  the  vertical  line  through  A. 


90 


MECHANICS 


[AKT.  61 


attached  at  B  and  held  at  A  and  by  a  second  cord  AC.  Find  the  tension  in 
each  cord  and  the  reaction  of  the  platform  when  the  bar  makes  an  angle 
of  35  degrees  with  the  horizontal  and  both  cords  are  horizontal. 

Ans.  P  =  Q  =  17.14  lb.;  R  =  28  Ib. 

11.  Solve  Problem  10  if  AC  is  horizontal,  BD  makes  an  angle  of  15  de- 
grees with  the  horizontal,  and  the  point  D  is  higher  than  B. 

Ans.     Q  =  12.83  lb.;  P  =  12.39  lb.;  R  =  24.68  lb. 

12.  Solve  Problem  10  if  AC  is  horizontal,  BD  makes  an  angle  of  15  degrees 
with  the  horizontal,  and  the  point  D  is  lower  than  B. 

13.  Solve  Problem  10  if  BD  is  perpendicular  to  AB,  and  also,  if  BD  is 
vertical. 

14.  In  Fig.  89,  the  cords  AC  and  BD  are  horizontal  and  the  force  P  is  10 
pounds.     Find  the  angle  which  the  bar  makes  with  the  horizontal. 

15.  A  bar  AB  is  10  feet  long,  weighs  30  pounds,  and  has  its  center  of  mass 
4  feet  from  A.     The  end  A  rests  on  a  horizontal  floor  and  the  end  B  rests 
against  a  vertical  wall.     Both  ends  are  so  constructed  as  to  have  practically 
no  friction.     The  bar  is  held  by  a  cord  attached  1  foot  from  A.     Find  the 

tension  in  the  cord  and  the  reactions  at  the  floor  and 
at  the  wall  when  the  bar  makes  an  angle  of  30  de- 
grees with  the  vertical  and  the  cord  is  horizontal. 
In  what  respect  does  this  differ  from  Problem  10? 

16.  Solve  Problem  15  if  the  wall  makes  an  angle 
of  15  degrees  with  the  vertical,  away  from  the  bar, 
as  shown  in  Fig.  90. 


FIG.  91. 


In  order  to  eliminate  two  unknowns,  it  is  advisable  to  take  moments 
about  the  point  of  intersection  of  the  vertical  line  through  A  with  the  line 
of  the  horizontal  cord.  It  is  best  to  use  the  horizontal  and  vertical  com- 
ponents of  Q  in  this  moment  equation.  A  part  of  the  solution  is 

30  X  4  sin  30°  =  Q(9  cos  30°  cos  15°  -f  10  sin  30°  sin  15°), 
Q  =  6.800  lb, 
P  =  Q  cos  15°  =  6.569  lb. 

The  student  will  verify  these  results  and  check  by  moments  about  A.  He 
should  find  R  by  vertical  resolution  and  check  by  moments  about  B. 

17.  The  bar  of  Problem  15  is  placed  upon  two  inclined  planes,  Fig.  91. 
The  end  A  rests  on  a  plane  which  makes  an  angle  of  65  degrees  to  the  left  of 
the  vertical,  and  the  end  B  rests  on  a  plane  which  makes  an  angle  of  50 
degrees  to  the  right  of  the  vertical.  Find  the  reaction  of  each  plane  and 
the  angle  which  the  bar  makes  with  the  horizontal. 

Compute  the  reactions  by  resolving  parallel  to  the  inclined  planes.     Then 


CHAP.  V]    NON-CONCURRENT  CO-PLANAR  FORCES       91 


find  the  angle  by  moments  about  A.     It  is  best  to  use  the  components  of 
the  force  P  in  the  moment  equation. 

30  X  4  cos  6  =  P  sin  40°  X  10  sin  0  +  P  cos  40°  X  10  cos  6, 
12  =  P  (sin  40°  tan  0  +  cos  40°). 

Check  by  moments  about  B.  Ans.     9  =  8°  07'. 

62.  Condition  for  Independent  Equations. — //  two  moment 
equations  of  equilibrium  are  written  for  a  problem  of  non-concur- 
rent, coplanar  forces,  and  these  equations  are  combined  into  a  sin- 
gle equation,  the  resulting  equation  is  equivalent  to  a  resolution 
perpendicular  to  the  line  joining  the  two  origins  of  moment. 

In  Fig.  92,  A  and  B  are  taken  as  the  two  origins  of  moment. 
The  point  B  is  the  origin  of  coordinates  and  the^X"  axis  is  passed 
through  the  point  A .  The  distance  between  A  and  B  is  equal  to 
c.  A  force  PI  at  an  angle  ai  with  the  X  axis  passes  through  the 


X 


/•»  B 

FIG.  92. 

point  (xi,  yi)  and  a  force  P2  at  an  angle  a2  with  the  -X"  axis  passes 
through  the  point  (x2}  yz),  etc.  Taking  moments  about  B, 

XiPi  sin  ai  —  2/i-Pi  cos  «i  +  XzPz  sin  a2  —  2/2^*2  cos  az  +  etc. 
=  0.  (1) 

Taking  moments  about  A} 

(xi  +  c)Pi  sin  ai—  yfi  cos  «i  +  (xz  +  c)P2  sin  a2  —  yzPz  cos 
a2  +  etc  =  0.  (2) 

Subtracting  Equation  (1)  from  Equation  (2), 

c(Pi  sin  «i  +  P2  sin  a2  +  etc.)  =  0.  (3) 

The  term  in  the  parenthesis  of  Equation  (3)  is  the  sum  of  the 
components  parallel  to  the  Y  axis;  therefore,  Equation  (3)  proves 
the  proposition. 

If  two  moment  equations  are  written  and  one  resolution  equa- 
tion, and  if  the  resolution  is  taken  along  a  direction  which  is 


92  MECHANICS  [ABT.  62 

perpendicular  to  the  line  joining  the  two  origins  of  moment,  the 
three  equations  will  not  be  independent.  If  three  moment  equa- 
tions are  written,  and  if  the  three  origins  of  moment  lie  on  the 
same  straight  line,  the  three  equations  will  not  be  independent. 
For  concurrent,  coplanar  forces  in  equilibrium  there  are 
two  unknowns,  and  two  independent  equations  may  be  written. 
These  may  be: 

(1)  Two  resolutions, 

(2)  One  moment  and  one  resolution. 

(3)  Two  moments. 

For  non-concurrent,  coplanar  forces  there  are  three  unknowns, 
and  three  independent  equations  may  be  written.  These  are 
the  same  as  the  equations  for  concurrent  forces  with  the  addition 
of  one  moment  equation.  If  the  forces  are  all  parallel,  there  can 
be  only  one  independent  resolution  equation,  and  only  two  un- 
known quantities. 

When  one  resolution  and  one  moment  equation  are  written 
for  concurrent  forces,  the  resolution  must  not  be  taken  perpen- 
dicular to  the  line  which  joins  the  origin  of  moments  and  the 
point  of  application  of  the  forces.  When  one  resolution  and  two 
moment  equations  are  written  for  non-concurrent  forces,  the 
resolution  must  not  be  taken  perpendicular  to  the  line  which 
joins  the  two  origins  of  moment. 

When  two  moment  equations  are  written  for  concurrent  forces, 
the  two  origins  of  moment  and  the  point  of  application  of  the 
forces  must  not  lie  on  the  same  straight  line.  When  three 
moment  equations  are  written  for  non-concurrent  forces,  the 
three  origins  of  moment  must  not  lie  on  the  same  straight  line. 
63.  Direction  Condition  of  Equilibrium. — When  a  body  is  in 
equilibrium  under  the  action  of  three  non-concurrent  forces, 

in  order  that  any  one  of  these  forces 
may  lie  in  the  line  of  the  resultant 
of  the  other  two  forces,  it  is  necessary 
that  all  three  forces  intersect  at  a 
single  point.  In  Fig.  93,  the  bar 
AB  is  hinged  at  A  and  supported 
by  a  cord  at  B.  The  vertical  line 
through  the  center  of  mass  at  the  point  C  intersects  the  line  of 
the  cord  at  D.  In  order  that  the  bar  may  be  in  equilibrium,  the 
direction  of  the  hinge  reaction  must  be  such  that  its  line  of  action 
shall  pass  through  D.  This  relation  is  sometimes  called  the 


CHAP.  V]    NON-CONCURRENT  CO-PLANAR  FORCES       93 

geometrical  condition  of  equilibrium.     In  this  book  it  will  be 
called  the  direction  condition  of  equilibrium. 

Example  I 

A  horizontal  bar  5  feet  long,  weighing  30  pounds,  with  its  center  of  mass 
2  feet  from  the  left  end,  is  supported  by  cords  at  the  ends.  The  cord  at 
the  left  end  makes  an  angle  of  30°  to  the  left  of  the  vertical.  Find  the 
direction  of  the  cord  at  the  right  end. 

From  the  right-angled  triangles  of  Fig.  94, 

CD  =  2  tan  60°  =  3  tan  6, 

tan.  =2X137321  =  1.1547, 
6  =  49°  06'. 

The  cord  at  the  right  end  makes  an  angle 

of  40°  54'  to  the  right  of  the  vertical.     The 

magnitude  of  the  forces  may  be  found  by  t       pIG    94 

moments  or  resolutions. 

The  direction  condition  may  take  the  place  of  one  moment  equation  in 
the  solution  of  a  problem  of  non-concurrent  forces.  The  other  necessary 
equations  may  be  resolutions. 

This  method  of  solution  is  especially  valuable  when  only  the  direction 
is  required.  In  practical  work  it  is  often  desirable  to  know  the  direction 
of  a  force,  in  order  to  put  in  a  support  in  the  best  position.  The  engineer 
should  cultivate  the  habit  of  observing  the  direction  of  forces  in  actual 
structures. 

Problems 

1.  A  horizontal  beam,  12  feet  long,  with  its  center  of  mass  5  feet  from  the 
left  end,  is  supported  by  two  posts.     One  post  at  the  left  end  makes  an 
angle  of  25  degrees  to  the  right  of  the  vertical.     What  should  be  the  direc- 
tion of  the  post  at  the  right  end  in  order  that  the  compression  in  each  post 
shall  be  parallel  to  its  length?     Solve  also  if  the  second  post  is  2  feet  from 
the  right  end. 

2.  Solve  Problem  1  if  one  post  is  at  the  left  end  and  the  other  is  3  feet 
from  the  left  end. 

3.  A  derrick  boom  is  20  feet  long,  makes  an  angle  of  40  degrees  with  the 
horizontal,  and  carries  a  load  of  400  pounds.     The  mast  is  12  feet  long. 
One  guy  rope,  which  is  in  the  plane  of  the  mast  and  boom,  makes  an  angle 
of  30  degrees  with  the  horizontal.     Find  the  direction  of  the  reaction  at 
the  bottom  of  the  mast  by  means  of  the  direction  condition  of  equilibrium. 
With  this  direction  known,  draw  the  force  triangle  for  the  three  external 
forces,  which  are  the  weight  on  the  boom,  the  tension  of  the  guy  rope,  and 
the  reaction  at  the  bottom  of  the  mast.     Also  solve  completely  by  Bow's 
method  for  all  external  and  internal  forces.     Compare  the  two  diagrams. 

4.  A  bar  7  feet  long,  with  its  center  of  mass  3  feet  from  the  left  end,  is 
supported  by  a  rope  at  the  left  end  which  makes  an  angle  of  30  degrees  to 
the  left  of  the  vertical,  and  a  rope  at  the  right  end  which  makes  an  angle  of 
35  degrees  to  the  right  of  the  vertical.     What  angle  does  the  bar  make  with 


94  MECHANICS  [ART.  63 

the  horizontal?     If  the  bar  weighs  60  pounds,  what  is  the  tension  in  each 
rope? 

From  Fig.  95, 

DF  =  DE  +  EF. 
4  cos  6  tan  55°  =  3  cos  6  tan  60°  +  7  sin  6;  . 

tan0 


f5 

V 

£M 
TJ-J 

e 

~-^r~ 

V 

/' 

\ 

i 

\ 

\ 

601b.      , 

^ 

i  • 

/ 

\ 

/ 

\             / 

\           / 

\    i     / 

\ 
\ 

>'F 

FIG.  95. 

By  a  resolution  perpendicular  to  the  direction  of  the  rope  at  the  left  end, 

.  Q  cos  25°  =  60  sin  30°, 

Q  =  30  sec  25°  =  30  X  1.1034  =  33.10  Ib. 

Check  Q  and  6  by  moments  about  the  left  end.     Solve  for  P  by  a  resolution 
equation  and  check. 

6.  Find  the  direction  of  the  bar  in  Problem  17  of  Art.  61  by  means  of  the 
direction  condition  of  equilibrium.  T /" 

6.  A  bar  5  feet  long,  weighing  2#  pounds,  with  its  center  of  mass  2  feet 
from  one  end,  is  placed  across  the  inside  of  a  hollow  cylinder  which  is  6  feet 
in  diameter.  The  ends  of  the  bar  are  ffictionless.  Find  the  position  of 
equilibrium  and  the  normal  reactions  at  the  ends. 

Ans.     The  bar  makes  an  angle  of  16°  47'  with  the  horizontal. 

When  a  body  is  in  equilibrium  under  the  action  of  four  forces, 
the  resultant  of  two  of  these  forces  must  lie  in  the  line  of  the 
resultant  of  the  other  two  forces.  In  order  that  this  may  happen, 
the  resultant  of  two  of  these  forces  must  pass  through  the  inter- 
section of  the  lines  of  action  of  the  others. 

% 

Example  II 

A  ladder  20  feet  long,  weighing  40  pounds,  with  its  center  of  mass  8  feet 
from  the  lower  end,  rests  on  a  smooth  horizontal  floor  and  leans  against  a 


CHAP.  V]    NON-CONCURRENT  CO-PLANAR  FORCES       95 


smooth  vertical  wall.     It  is  held  from  slipping  by  a  horizontal  force  of  12 
pounds  at  the  bottom.     Find  the  position  of  equilibrium. 

The  resultant  of  the  horizontal  force  and  the  vertical  reaction  at  the 
bottom  must  pass  through  the  point  D, 
(Fig.  96),  at  which  the  vertical  line 
through  the  center  of  mass  intersects 
the  horizontal  line  through  the  top  of 
the  ladder.  By  vertical  resolution,  the 
vertical  reaction  at  the  bottom  is  found 
to  be  40  pounds.  The  resultant  of  40 
pounds  and  12  pounds  makes  an  angle 
with  the  vertical  whose  tangent  is  0.3. 
From  the  figure, 

DE  =  20  sin  6  =  8  cos  9  cot  a; 


tan  0  = 


0.3  X  20 
6  =  53°  08'. 


1.3333; 


FIG.  96. 


Problems 


7.  A  bar  10  feet  long,  weighing  40  pounds,  with  its  center  of  mass  at  the 
middle,  rests  on  a  smooth  horizontal  floor  and  leans  against  a  smooth  vertical 
wall.  It  is  held  by  a  horizontal  pull  of  10  pounds  applied  2  feet  from  the 
lower  end.  Find  the  reactions  at  the  floor  and  at  the  wall  by  resolutions, 
and  find  the  angle  which  the  bar  makes  with  the  vertical  by  means  of  the 
direction  condition  of  equilibrium.  Ans.  21°  48'  with  the  vertical. 

'  8.  A  bar  *j?  feet  long,  with  its  center  of  mass  5  feet  from  the  lower  end, 
rests  on  a  smooth  horizontal  floor  and  leans  against  a  wall  which  makes  an 
angle  of  15  degrees  with  the  vertical,  away  from  the  bar.  The  bar  makes 
an  angle  of  35  degrees  with  the  vertical.  It  is  held  by  a  horizontal  force 
applied  2  feet  from  the  bottom.  Determine  the  three  unknown  forces 
graphically,  using  the  direction  condition  of  equilibrium  to  find  the  angles. 

64.  Trusses. — A  truss  is  a  jointed  frame.  Since  a  triangle  is 
the  only  jointed  frame  which  retains  its  form  when  loaded,  a 
truss  is  made  up  of  a  connected  series  of  triangular  elements. 


FIG.  97. 


Figure  97  shows  a  truss  which  is  entirely  supported  from  one  end. 
A  truss  supported  at  one  end  is  called  a  cantilever  truss.     The  dis- 


96 


MECHANICS 


[ABT.  64 


tances  AB,  BC,  and  CD,  are  panel  lengths.  The  top  of  the 
truss  is  called  the  top  chord.  The  lower  members,  EF  and  FG, 
comprise  the  bottom  chord.  Either  chord  may  be  one,  continuous, 
rigid  body,  but  the  calculations  are  made  on  the  assumption  that 
each  panel  is  a  separate  body  which  is  pin-connected  to  the  re- 
mainder of  the  truss. 

When  loads  are  applied  at  the  joints  of  a  truss,  the  internal 
stresses  are  calculated  as  a  series  of  problems  of  concurrent  forces. 
Bow's  method  is  used  for  the  graphical  solution.  The  truss  of 
Fig.  97  may  be  calculated  in  this  way,  beginning  at  the  right  end, 
where  there  are  only  two  unknowns. 

In  most  trusses,  no  point  can  be  found  at  which  there  are  only 
two  unknowns.  It  is  then  necessary  to  find  first  the  external 
reactions  as  a  problem  of  non-concurrent  forces.  After  these 
reactions  have  been  found,  the  forces  at  the  separate  joints  are 
calculated  as  a  series  of  connected  members. 

Example 

The  truss  shown  in  the  diagram  of  Fig.  98  is  supported  at  the  left  end  on 
rollers,  which  make  the  reaction  vertical.  It  is  hinged  at  the  right  end. 


,  Hinge 

H 


The  reaction  at  the  right  end  has  a  horizontal  component  H  and  a  vertical 
component  R2.  Solve  for  the  reactions  and  then  solve  for  all  the  internal 
forces. 

Moments  about  the  left  support  eliminate  Ri  and  H. 

4200  X  15  =  63000 
4000  X  21  =  84000 
2800  X  33  =  92400 


42#2  =  239400 
R2  =      5700  lb. 


It  is  best  to  use  the  components  of  the  load  of  4200  pounds  when  moments 


CHAP.  V]     NON-CONCURRENT  CO-PLANAR  FORCES       97 


are  taken  about  the  hinge.  The  end  posts  are  15  feet  in  length.  Sin  0  = 
ye  =  0.8000;  cos  0  =  0.6000.  The  force  of  4200  pounds  at  right  angles  to 

the  end  post  makes  an  angle  6  with  the  horizontal.  The  horizontal  com- 
ponent of  4200  pounds  is  4200  X  0.8  =  3360  pounds.  The  vertical  com- 
ponent is  4200  X  0.6  =  2520  pounds.  Using  these  components  instead  of 
the  4200-pounds  force,  and  taking  moments  about  the  right  end  of  the  truss, 

2800  X  9  =  25200 
4000  X  21  =  84000 
2520  X  33  =  83160 


192360  counter-clockwise, 
3360  X  12  =  40320  clockwise, 
42tfi  =152040 

R!  =      3620  Ib. 
Resolve  vertically  for  a  check, 


5700 
3620 


2520 
4000 
2800 


9320   =  9320 

The  horizontal  component  of  the  hinge  reaction  is  obtained  by  a  horizontal 
resolution, 

H  =  3360  Ib. 

This  problem  of  non-concurrent,  coplanar  forces  has  now  been  solved  by 
two  moments  and  one  resolution  and  partly  checked  by  a  second  resolution, 
It  might  well  be  checked  again  by  moments  about  the  point  of  application 
of  the  4200-pound  load,  or  about  the  middle  of  the  top  chord. 


2520# 


4000# 


3800^' 


3360# 


It  is  not  best  to  use  the  components  of  the  inclined  force  in  both  moment 
equations,  for,  if  an  error  is  made  in  the  computation,  the  reactions  will 
check  and  still  be  incorrect. 

Figure  99  shows  the  reactions  for  the  truss  of  Fig.  98.  The  force  of  4200 
pounds  is  replaced  by  its  components.  If  a  force  and  its  components  are 
written  on  the  same  diagram,  one  or  the  other  should  be  enclosed  in  a 
7 


98  MECHANICS  [ART.  64 

parenthesis,  or  otherwise  marked,  on  account  of  the  danger  of  using  both 
the  force  and  its  components  in  the  same  equation. 

With  the  external  forces  known,  the  internal  forces  may  now  be  computed. 
The  algebraic  method  of  solution,  joint  at  a  time,  will  be  given  first.  This 
will  be  followed  by  the  graphical  solution.  Figure  99  has  been  lettered  by 
Bow's  method.  Each  force  will  be  represented  by  two  letters.  (It  is  most 
convenient  in  the  algebraic  method  to  use  a  single  letter  for  each  force. 
The  two  letters,  will  be  used  here,  however,  in  order  to  compare  the  results 
more  readily  with  the  graphical  solution.) 

Beginning  with  joint  No.  1  as  the  free  body  and  resolving  vertically, 

ag  sin  0  =  3620, 

comPress*on- 


Resolving  horizontally, 

fff  =  ag  cos  e  =  4525  X  0.6,  gf  =  2715  Ib.  tension. 

The  tension  in  GF  might  have  been  calculated  by  moments  about  joint  No.  2, 
gf  X  12  =  3620  X  9. 

At  joint  No.  2,  AG  pushes  upward.  Its  horizontal  component  at  the  top 
is  the  same  as  at  the  bottom,  or  2715  pounds.  Its  vertical  component  is 
3620  pounds.  Resolving  horizontally, 

bh  =  3360  +  2715,  bh  =  6075  Ib.  compression. 

Resolving  vertically,  and  assuming  that  gh  pulls  downward, 

ag  sin  6  =  2520  +  gh, 

gh  =  3620  -  2520,  gh  =  1100  Ib.  tension. 

Resolving  vertically  at  joint  No.  3, 

Ai  sin  45°  =  gh  =  1100, 

hi  =  1100  X  1.4142,  hi  =  1556  Ib.  compression. 

Resolving  horizontally,  HI  pushes  toward  the  left  and  GF  pulls  toward  the 
left,  consequently  IF  must  pull  toward  the  right.  The  horizontal  compo- 
nent of  the  force  in  HI  is  the  same  as  the  vertical  component; 

if  =  2715  +  1100,  if  =  3815  Ib.  tension. 

Resolving  vertically  at  joint  No.  4, 
ij  sin  45°  =  4000  -  1100  =  2900, 

ij  =  2900  X  1.4142,  ij  =  3701  Ib.  compression. 

Resolving  horizontally,  assuming  that  cj  is  compression, 

cj  =  bh  +  hi  cos  45°  —  ij  cos  45°, 

cj  =  6075  +  1100  -  2900,  cj  ''=  4275  Ib,  compression. 

At  joint  No.  5, 

kd  cos  6  =  cj  =  4275, 

4275 

^  =  "Oil"'  ^  =  ^^  ***'  compression. 

jk  +  2800  =  Kd  sin  6  =  7125  X  0.8, 
jk  =  5700  -  2800,  jk  =  2900  Ib.  tension. 


CHAP.  V]     NON-CONCURRENT  CO-PLANAR  FORCES       99 


At  joint  No.  6, 


jk  =  ij  sin  45°  =  2900  Ib. 

kf  =  if-  ij  cos  45°  =  3815  -  2900, 


kf  =  916  Ib.  tension. 


At  joint  No.  7, 


kd  sin  0  =  fe  =  Rz 
R2  =  7125  X0.8  =  5700  Ib. 

kd  cos  6  —  kf  =  de  =  H, 
H  =  4275  -  915  =  3360  Ib. 

These  last  two  results  check  the  hinge  reaction.     The  values  of  jk  from  joints 
5  and  6  also  check. 

Figure  100  is  the  graphical  solution  for  Fig.  98.     The  diagram 
begins  with  a/,  the  vertical  reaction   at  the  left  end.     The  last 


FIG.  100. 

point  is  k.  This  point  should  fall  on  the  horizontal  line  igf, 
in  order  that  the  line  kf  may  be  horizontal.  On  account  of  errors 
in  the  drawing,  the  point  k  is  slightly  above  igf.  The  distance 
from  k  to  the  horizontal  line  is  the  closing  error. 

The  external  forces  fa,  ab,  be,  cd,  de,  and  ef  form  a  polygon. 
A  line  representing  an  external  force  is  recognized  by  the  fact 


100  MECHANICS  [ART.  65 

that  it  carries  only  one  arrow.     The  external  fdrce  polygon  also 
must  close. 

Problems 

1.  Draw  the  space  diagram,  Fig.  98,  to  the  scale  of  1  inch  =  8  feet,  or 
1  inch  =  10  feet.     With  the  external  reactions  known  from  the  above 
example,  draw  the  force  diagram  to  the  scale  of  1  inch  =  1000  pounds. 
Draw  all  forces  parallel  to  the  corresponding  lines  on  the  space  diagram. 
Measure  all  lengths  on  the  force  diagram  and  put  the  values  in  pounds  on 
the  space  diagram. 

2.  Solve  the  above  Example  by  Bow's  method,  putting  /  at  the  top  and 
a  at  the  bottom  of  the  first  line  a/.  c 

3.  In  Fig.  98,  make  the  load  in  AB  2800  pounds,  the  load  in  BC  3200 
pounds,  and  the  load  in  CD  3500  pounds.     Find  the  reactions  and  check. 
Solve  for  all  internal  forces,  joint  at  a  time.     Solve  by  Bow's  method  to 
the  same  scale  as  Problem  1. 

It  is  a  goopl  plan  to  carry  Bow's  method  along  with  the  algebraic 
solution  for  the  internal  forces.  Calculate  the  forces  at  a  given 
joint,  and  then  draw  the  part  of  the  force  diagram  for  that  joint. 


800#  600#    I500# 

& 

90' 


3\       I      I 


FIG.  101. 

4.  Figure  101  shows  a  roof  truss  hinged  at  the  right  end.  Find  the  reac- 
tions and  check.  Find  all  the  internal  forces  joint  at  a  time.  Construct 
a  space  diagram  to  the  scale  of  1  inch  =  8  feet  or  1  inch  =  10  feet.  Letter 
and  solve  by  Bow's  method  to  the  scale  of  1  inch  =  500  pounds. 

65.  The  Method  of  Sections. — It  is  often  desirable  to  find  the 
stress  in  a  few  members  of  a  truss  or  other  structure.  For  this 
purpose,  the  truss  may  be  imagined  to  be  cut  at  some  surface 
into  two  portions.  Either  of  these  portions  may  be  regarded 
as  a  free  body  in  equilibrium  under  the  action  of  the  external 
forces  on  its  side  of  the  surface,  and  of  the  internal  forces  which 
cross  the  surface  from  the  other  portion.  These  internal  forces 
which  cross  the  imaginary  surface  are  external  to  the  portion 
under  consideration.  If  there  are  not  more  than  three  unknown 


CHAP.  V]    NON-CONCURRENT  CO-PLANAR  FORCES      101 

forces  acting  on  the  portion  in  question,  these  may  be  deter- 
mined as  a  problem  of  non-concurrent  forces. 

Example 

Figure  102  represents  the  truss  of  Fig.  99.  The  external  reactions  are  known. 
The  truss  may  be  regarded  as  divided  at  the  surface  SR,  which  cuts  the 
members  CJ,  JI,  and  IF.  The  portion  to  the  right  of  SR  will  be  taken  as 
the  free  body  This  portion  is  shown  separately  in  Fig.  102,  II.  The 
forces  which  act  on  the  free  body  are  the  known  forces  of  2800  pounds, 
3360  pounds,  and  5700  pounds,  together  with  the  unknown  forces  in  the 
members  C/,  ,77,  and  IF.  The  direction  of  each  of  these  members  is  known; 
hence,  the  unknowns  are  the  magnitudes  of  the  three  forces. 


E5ZO 


36ZO 


FIG.  102. 


The  members  «//  and  IF  intersect  at  joint  No.  6.     If  moments  are  taken 
about  this  joint,  the  only  unknown  is  the  force  in  CJ. 


12  X  cj  =  9  X  5700, 


cj  =  4275  Ib. 


The  reaction  of  5700  pounds  tends  to  turn  the  portion  of  the  truss  in  a 
counter-clockwise  direction  about  joint  No.  6.  The  force  in  CJ  must  turn 
in  the  opposite  direction.  Consequently  CJ  must  push  toward  the  right. 

cj  =  4275  Ib.  compression. 

The  members  JI  and  CJ  intersect  at  joint  No.  4,  Fig.  102,  I.  If  moments 
are  taken  about  this  joint,  the  only  unknown  is  the  force  in  IF.  (This 
origin  of  moments  is  outside  the  portion  under  consideration.  If  a  body  ia 


102 


MECHANICS 


[AKT.  65 


in  equilibrium,  the  sum  of  the  moments  is  zero  for  any  origin  whatever, 
whether  inside  the  body  or  entirely  away  from  it.) 


2800  X  12 
3360  X  12 


=   33600 
=    40320 


73920  clockwise, 
5700  X  21    =  119700    counter-clockwise 


if  X  12  =    45780, 

if  =  3817  Ib.  tension. 

Two  of  the  unknowns  have  now  been  found  by  two  moment  equations. 
The  third  unknown  is  best  found  by  a  vertical  resolution, 

ij  sin  45°  =  5700  -  2800  =  2900, 
ij  =  2900  cosec  45°  =  2900  X  1.4142, 

ij  =  3701  Ib.  compression. 

It  is  not  at  all  necessary  to  make  a  separate  drawing  of  the  portion  of  the 
truss  which  is  taken  as  the  free  body.  This  has  been  done  in  Fig.  102,  II, 
in  order  to  simplify  the  explanation.  Generally  the  original  space  diagram, 
Fig.  102,  I,  is  employed  and  a  section  line  is  drawn  across  to  indicate  the 
boundary  of  the  portion  in  equilibrium. 

The  solution  of  a  problem,  joint  at  a  time,  may  be  regarded 
as  a  special  case  of  the  method  of  sections.  A  single  joint  is  cut 
off  as  the  free  body  and  the  forces  are  concurrent.  The  term, 
"method  of  sections"  is  applied  to  cases  like  that  of  Fig.  102,  II, 
in  which  the  forces  on  the  free  body  are  non-concurrent. 


600# 


800* 


1000' 


Problems 

1.  In  Fig.  102,  use  the  portion  of  the  truss  to  the  left  of  the  section  RS 
as  the  free  body  and  solve  for  the  forces  in  CJ,  JI,  and  IF.     Take  moments 
about  the  same  points  as  in  the  example  above. 

2.  In  Fig.  102,  make  a  section  VU,  which  cuts  the  members  BH,  HI, 
and  IF,  and  find  the  force  in  the  members  cut  without  using  any  other 

internal  forces.  Solve  first  with  the 
portion  to  the  left  of  the  section  as 
the  free  body  and  check  with  the 
portion  to  the  right  of  the  section  as 
the  free  body. 

3.  Make  a  section  WX  in  Fig.  102, 
separating  the  end  post  AG  as  a  free 
body.     Find  the  force  in  HG  by  a 
vertical  resolution. 

4.  In  Problem  3  of  Art.  64,  cal- 
culate the  external  reactions,  and 

then  find  bh  and  hi  by  sections  without  using  any  other  internal  forces. 

6.  In  Fig.  103,  find  the  reactions  and  check.  Make  a  section  through 
BH,  HG,  and  OF  and  find  the  forces  in  these  members,  without  using  any 
other  internal  forces. 


10 


F 
FIG.  103. 


CHAP.  V]    NON-CONCURRENT  CO-PLANAR  FORCES      103 


Could  you  solve  Problem  5  by  means  of  a  section  through  BH,  HI,  IJ, 
and  JF1     Why? 

6.  Solve  Problem  5  for  all  internal  forces,  joint  at  a  time. 

7.  Solve  Problem  5  for  all  internal  forces  by  Bow's  method  to  the  scale 
of  1  inch  =  400  pounds. 

8.  In  Fig.  104,  the  loads  in  AB  \         B        \ 
and    BC   are    each    1200   pounds. 

Find  the  reactions.  Find  bf  and 
fd  by  sections.  Find  all  internal 
forces,  joint  at  a  time. 

9.  In  Fig.  104,  the  load  in  AB  is 
800  pounds,  and  the  load  in  BC  is   i 
1200  pounds.     Find  the  reactions.    ' 
Find  bf  and  fd  by  sections.     Check 
by  a  horizontal  resolution. 

10.  In  Fig.  105,  find  the  reactions  in  BA,  AF,  and  FE.     Check.     Find 
ci,  ij,  and  je  by  sections.     Begin  at  the  right  end  and  solve  for  all  forces, 
joint  at  a  time.     Begin  at  the  right  end  and  solve  graphically. 


*5K. 


1600* 


i  a 


£• 
FIG.  105. 

66.  Jointed   Frame    with   Non-concurrent   Forces. — In    the 

jointed  frames  considered,  the  loads  have  been  applied  at  the 
joints;  consequently  the  intervening  members  transmitted  force 
only  in  the  direction  of  their  length.  A  member  to  which  only 
two  forces  are  applied,  is  called  a  two-force  member.  If  the  forces 
are  in  equilibrium,  they  must  lie  along  the  same  straight  line,  and 
if  they  are  applied  at  opposite  ends  of  the  member,  the  direction 
of  each  force  must  be  in  the  line  of  the  member.  A  member  to 
which  three  forces  are  applied  is  called  a  three-force  member. 
It  is  not  necessary  that  any  of  these  forces  be  in  the  direction  of 
its  length. 

Example 

Figure  106  shows  two  links  which  are  connected  to  each  other  and  to  the 
supports  by  smooth  hinges.  The  left  link  carries  a  load  of  30  pounds  1  foot 
from  the  left  end;  the  right  link  carries  a  load  of  50  pounds  at  the  middle. 
The  problem  is  to  find  the  reactions  of  the  supporting  hinges  and  the  in- 
ternal reaction  at  the  hinge  which  connects  the  links. 

Equilibrium  equations  may  be  written  for  the  two  links  separately,  or 


104 


MECHANICS 


[ART.  66 


for  both  together  as  a  single  free  body.  It  is  generally  advisable  to  write 
first  the  equations  for  the  entire  structure,  and  calculate  as  many  external 
forces  as  possible;  then  write  equations  for  parts  of  the  structure  to  obtain 
the  remainder. 

The  links  and  their  support  form  a  3 : 4 : 5  triangle.  Cos  6  =  0.6;  sin  0  = 
0.8;  cos  <£  =  0.8;  sin  <f>  =  0.6. 

The  vertical  component  of  the  reaction  at  the  left  hinge  is  Vi.  The 
vertical  component  of  the  reaction  at  the  right  hinge  is  F2.  The  horizontal 
component  at  the  left  hinge  is  H i.  The  horizontal  component  at  the  right 
hinge  is  H2. 


FIG.  106. 

Taking  moments  about  the  left  hinge, 

5F2  =  30  X  0.6  +  50  X  3.4, 
30  X  0.6  =     18 
50  X  3.4  =  170 
5F2  =  188 
Taking  moments  about  the  right  hinge, 

50  X  1.6  =    80 
30  X  4.4  =  132 


V,  =  37.6  Ib 


5Vi  =  212  Vi   =  42.4  Ib. 

There  are  still  two  external  forces,  HI  and  H2.  By  horizontal  resolution 
these  are  known  to  be  equal  and  opposite.  All  the  possible  equations  for 
a  problem  of  non-concurrent  forces  have  now  been  used  and  there  is  still 
one  unknown  to  be  found.  The  vertical  reactions  depend  entirely  upon  the 
horizontal  distances  of  the  lines  of  the  loads  from  the  two  supports  and  are 
independent  of  the  length  of  the  links  and  the  kind  of  connections  between 
them. 

To  find  H2,  the  right  link  may  be  taken  as  the  free  body  and  a  moment 
equation  written  about  the  hinge  which  connects  the  two  links. 
37.6  X  3.2  =  120.32  counter-clockwise, 

50 X  1.6  =    80 . 00  clockwise, 

2.4#2  =      40.32,  H2  =  16.8  Ib. 

The  horizontal  reaction  at  the  left  hinge  is  16.8  pounds  toward  the  left. 


CHAP.  V]     NON-CONCURRENT  CO-PLANAR  FORCES      105 


In  Fig.  106,  Hz  is  the  horizontal  component  of  the  force  from  the  right 
link  to  the  left  link,  and  F3  is  the  vertical  component.  Equal  and  opposite 
forces  act  from  the  left  link  to  the  right  link. 

Using  the  left  link  as  the  free  body  and  taking  vertical  resolutions, 

yl  +  Vz  =  30  Ib. 

F3  =  30  -  42.4,    V3  =  12.4  Ib.  downward. 

If  F4  is  the  vertical  component  of  the  reaction  of  the  left  link  on  the  right 
link,  a  vertical  resolution  with  the  right  link  as  the  free  body  gives  an  equal 
force  in  the  opposite  direction. 

By  horizontal  resolutions,  with  either  link  as  the  free  body,  H3  =  16.8  Ib. 
Both  Hz  and  F3  may  be  checked  by  moments  about  the  left  hinge  with  the 
left  link  as  the  free  body. 

Problems 

1.  In  Fig.  107,  find  the  direction  and  magnitude  of  the  reactions  at  the 
supporting  hinges,  and  the  direction  and  magnitude  of  the  force  from  the 
inclined  member  to  the  horizontal  member  at  the  hinge  which  connects 
them. 

Ans.     Hl  =  #2  =  #3  =  163.92  Ib.; 

Fi  =  25.36  Ib.;  F2  =  134.65  Ib.;  F3  =  34.64  Ib.; 
fa  =  165.9  Ib.  left  8°  48'  up; 
,        JK2  =  212.1  Ib.  right  39°  24'  up; 
#3  =  167.5  Ib.  right  11°  32'  up. 


60 

Lr-  *'-  

-1 

Ib. 

ffl^,-                                  nr      \p/                                ^X». 

« 


./I 


K 


'6O 


V- 


FIG.  107. 


2.  In  Problem  1,  find  the  location  of  the  resultant  of  100  pounds  and  60 
pounds  by  moments.     Then,  with  the  direction  of  Ri  known  from  the  alge- 
braic solution,  find  the  direction  of  R2  by  means  of  the  direction  condition 
of  equilibrium  of  three  forces.     Finally,  draw  the  force  triangle  and  get 
the  magnitudes  of  Ri  and  #2. 

3.  In  Fig.  108,  find  the  direction  and  magnitude  of  the  reaction  at  each 
hinge. 

A  problem  in  which  the  members  of  a  connected  system  are 
subjected   to   non-concurrent   forces   may   be   changed  into   a 


106 


MECHANICS 


[ABT.  66 


problem  of  the  ordinary  type  by  considering  the  forces  between 
the  joints  as  replaced  by  forces  at  the  joints.  In  Fig.  109,  which 
is  the  same  as  Fig.  106,  the  load  of  50  pounds  at  the  middle  of  the 


200  Ib. 


FIG.  108. 


right  link  has  been  replaced  by  a  load  of  25  pounds  at  each  end  of 
the  member;  and  the  load  of  30  pounds  at  the  third  point  of  the 
left  link  has  been  replaced  by  20  pounds  at  the  left  end  and  10 


f- 

I    M 


II 


pounds  at  the  right  end.  The  load  of  10  pounds  at  the  right 
end  of  the  left  link  and  the  load  of  25  pounds  at  the  left  end  of  the 
right  link  together  are  equivalent  to  35  pounds.  As  far  as  the 


CHAP.  V]    NON-CONCURRENT  CO-PLANAR  FORCES      107 

external  forces  are  concerned,  the  problem  of  Fig.  109,  I,  is 
equivalent  to  the  problem  of  Fig.  106. 

In  Fig.  109,  I,  single  reactions  have  been  drawn  at  each  support,  instead 
of  the  horizontal  and  vertical  components.  These  components,  however, 
might  have  been  used.  . 

Figure  1 09,  II,  shows  the  graphical  solution.  At  the  lower  joint,  there  are 
two  unknowns  in  the  direction  of  the  links.  Beginning  with  the  known 
load  of  35  pounds  vertically  downward,  draw  the  force  triangle  dca.  At  the 
right  hinge,  the  unknowns  are  the  direction  and  magnitude  of  the  hinge 
reaction.  Lay  off  the  line  cb,  25  units  in  length,  in  a  vertical  direction. 
The  closing  line  ba  gives  the  direction  and  magnitude  of  the  hinge  reaction. 
The  horizontal  component  of  the  hinge  reaction  is  the  force  H2  of  Fig.  106, 
and  the  vertical  component  is  the  force  Vz>  The  left  reaction  de  is  found 
in  a  similar  manner. 

To  find  the  direction  and  magnitude  of  the  reaction  at  the  lower  hinge, 
the  load  of  25  pounds  is  regarded  as  acting  on  the  right  link,  and  the  load 
of  10  as  acting  on  the  left  link.  The  pin  may  be  regarded  as  replaced  by  a 
short  link  and  an  additional  letter  inserted  in  the  space  diagram.  On  the 
force  diagram  measure  10  units  downward  from  d  to  the  point/.  The  broken 
line  af  is  the  reaction  at  the  pin. 

Problems 

4.  In   Problem    1,   transfer  the  vertical  loads  to  the  joints  and  solve 
graphically. 

5.  In  Problem  3,  Fig.  108,  transfer  the  vertical  loads  to  the  joints  and 
solve  graphically. 

67.  Summary. — The  magnitude  and  direction  of  the  resultant 
of  a  set  of  non-concurrent,  coplanar  forces  is  the  vector  sum  of 
the  forces.  If  the  forces  are  all  parallel,  the  vector  sum  is  the 
algebraic  sum.  In  these  respects,  non-concurrent  forces  and 
concurrent  forces  are  alike. 

The  location  of  the  resultant  is  found  from  the  condition  that 
the  moment  of  the  resultant  about  any  point  is  equal  to  the 
moment  of  the  separate  forces  about  that  point. 

For  equilibrium,  the  force  polygon  must  close,  and  the  sum  of 
the  moments  about  any  point  must  be  zero. 

For  the  algebraic  solution  of  a  problem  of  the  equilibrium  of 
non-concurrent,  coplanar  forces,  three  independent  equations 
are  written.  These  may  be: 

1.  One  moment  equation  and  two  resolution  equations. 

2.  Two  moment  equations  and  one  resolution  equation. 

3.  Three  moment  equations. 


108  MECHANICS  [ART.  68 

When  two  moment  and  one  resolution  equation  are  written, 
the  resolution  must  be  not  taken  perendicular  to  the  line  which 
joins  the  two  origins  of  moment.  When  three  moment  equations 
are  written,  the  three  origins  of  moment  must  not  lie  in  the  same 
straight  line. 

When  the  forces  are  all  parallel,  there  can  be  only  two  unknowns 
and  two  independent  equations.  One  of  these  equations  must 
be  a  moment  equation. 

The  direction  condition  of  equilibrium  for  three  forces  requires 
that  the  lines  of  action  of  the  forces  must  meet  at  a  point.  The 
direction  condition  for  four  forces  is  that  the  resultant  of  two 
of  the  forces  must  pass  through  the  intersection  of  the  other  two. 
A  direction  condition  may  replace  a  moment  equation  in  the 
solution  of  a  problem  of  equilibrium.  The  direction  conditions  of 
equilibrium  are  especially  useful  in  problems  in  which  angles  are 
to  be  found. 

A  resolution  perpendicular  to  the  direction  of  an  unknown  force 
eliminates  that  force.  A  moment  equation  with  respect  to 
a  point  in  the  line  of  action  of  an  unknown  force  eliminates  that 
force.  It  is  often  desirable  to  take  moments  about  the  point  of 
intersection  of  two  unknown  forces,  or  about  a  point  in  the  line  of 
action  of  a  force  whose  direction  and  magnitude  are  both 
unknown. 

In  a  connected  system,  the  entire  system  is  first  treated  as  the 
free  body  and  as  many  of  the  reactions  as  possible  are  calculated. 
The  parts  of  the  system  are  then  treated  as  free  bodies.  When 
the  free  body  is  a  single  joint,  the  forces  are  concurrent.  The 
method  of  sections  divides  the  system  into  two  portions.  Either 
portion  may  be  treated  as  the  free  body  in  equilibrium.  The 
forces  which  act  on  the  portion  are  the  external  forces  on  its  side 
of  the  section,  and  the  internal  forces  in  the  members  which  are 
cut  by  the  section. 

68.  Miscellaneous  Problems 

1.  Two  bodies,  one  of  which  weighs  20  pounds  and  the  other  50  pounds, 
are  attached  to  the  ends  of  a  rope  which  runs  over  a  smooth  pulley.     The 
50-pound  mass  is  in  contact  with  a  smooth  plane  which  makes  an  angle  of 
15  degrees  with  the  horizontal.     Find  the  position  of  equilibrium,  the  reac- 
tion of  the  plane,  and  the  direction  and  magnitude  of  the  resultant  force  on 
the  pulley. 

2.  In  Problem  1,  what  is  the  maximum  angle  of  the  plane  with  the  hori- 
zontal in  order  that  equilibrium  may  be  possible? 


CHAP.  V]    NON-CONCURRENT  CO-PLANAR  FORCES      109 

3.  In  Problem  1,  both  bodies  are  ki  contact  with  the  plane.     The  per- 
pendicular distance  from  the  pulley  to  the  plane  is  5  feet.     Neglecting  the 
dimensions  of  the  pulley,  find  the  minimum  length  of  rope  for  possible 
equilibrium. 

4.  A  20-pound  mass  is  supported  by  two  cords.     One  cord  is  5  feet  long 
and  has  the  upper  end  attached  to  a  point  A.     The  second  cord  runs  over 
a  smooth  pulley  at  the  same  level  as  the  point  At  and  5  feet  therefrom,  and 
carries  a  load  of  12  pounds.     Find  the  position  of  equilibrium,  the  tension 
in  the  first  cord,  and  the  resultant  force  on  the  pulley.     There  are  two  possi- 
ble solutions  of  the  equations.     What  change  must  be  made  in  the  construc- 
tion of  the  apparatus  in  order  that  both  solutions  may  be  mechanically 
possible? 

5.  Solve  Problem  4  if  the  pulley  is  5  feet  from  A  along  a  line  which  makes 
an  angle  of  10  degrees  with  the  horizontal.     Write  moments  about  A  and 
solve  by  trial  and  error. 

Ans.     The  cord  from  A  makes  an  angle  of  29°  18'  with  the  vertical. 

6.  A  bar  10  feet  long,  weighing  60  pounds,  with  its  center  of  mass  6  feet 
from  the  left  end,  is  hinged  at  the  left  end  and  supported  by  means  of  a 
rope  at  the  right  end.     The  right  end  is  elevated  20  degrees  above  the 
horizontal.     The  rope  at  the  right  end  makes  an  angle  of  35  degrees  to  the 
left  of  the  vertical.     Find  the  tension  in  the  rope  and  the  direction  and  mag- 
nitude of  the  hinge  reaction  by  means  of  one  moment  equation  and  two 
resolution  equations.     Check  by  a  second  moment  equation. 

7.  Solve  Problem  6  by  means  of  the  direction  condition  >of  equilibrium. 
Compare  with  the  results  of  Problem  6. 

8.  A  box  is  5  feet  long  horizontally  and  3  feet  high.     It  weighs  240  pounds 
and  its  center  of  mass  is  at  the  center.     A  force  applied  at  the  upper  right 
edge  tips  the  box  about  the  lower  left  edge.     The  force  makes  an  angle  of 
25  degrees  to  the  left  of  the  vertical.     If  the  friction  is  sufficient  to  prevent 
sliding,  what  force  will  be  required  to  start  the  box? 

Ans.     103.4  Ib. 

9.  Solve  Problem  8  graphically  by  means  of  the  direction  condition  of 
equilibrium. 

10.  In  Problem  8,  what  is  the  direction  and  magnitude  of  the  minimum 
force  at  the  upper  right  edge  which  will  start  to  tip  the  box? 

11.  The  box  of  Problem  8  is  turned  through  an  angle  of  20  degrees.     Find 
the  direction  and  magnitude  of  the  smallest  force  at  the  upper  right  corner 
which  will  hold  it  in  this  position? 

12.  A  ladder  20  feet  long,  weighing  50  pounds,  with  its  center  of  mass 
8  feet  from  the  lower  end,  stands  on  a  smooth  horizontal  floor  and  leans 
against  a  smooth,  vertical  wall.     It  is  held  from  slipping  by  a  horizontal 
force  at  the  floor.     Find  this  force  and  the  reaction  at  the  floor  and  at  the 
wall  when  the  ladder  makes  an  angle  of  25  degrees  with  the  vertical. 

13.  The  ladder  of  Problem  12  stands  on  a  smooth  horizontal  floor  and 
leans  against  the  edge  of  a  wall  15  feet  in  height.     It  is  held  from  slipping  by 
a  horizontal  force  at  the  bottom.     Solve  for  the  unknowns  when  the  ladder 
makes  an  angle  of  30  degrees  with  the  horizontal. 

14.  Solve  Problem  12  if  the  horizontal  force  at  the  bottom  of  the  ladder 
is  16  pounds  and  the  position  is  unknown. 


110 


MECHANICS 


[ART.  68 


16.  Solve  Problem  12  if  the  horizontal  force  is  15  pounds  and  is  applied 
2  feet  from  the  bottom  of  the  ladder.  Check  your  results  by  means  of 
the  direction  .condition  of  equilibrium. 

*  16.  A  beam  20  feet  long,  weighing  80  pounds,  with  its  center  of  mass  at 
the  middle,  has  its  left  end  on  a  smooth  plane  which  makes  an  angle  of  60 
degrees  to  the  left  of  the  vertical,  and  its  right  end  on  a  smooth  plane  which 
makes  an  angle  of  70  degrees  to  the  right  of  the  vertical.  The  beam  carries 
a  load  of  60  pounds  5  feet  from  the  left  end,  and  a  load  of  100  pounds  4  feet 
from  the  right  end.  Find  the  reactions  of  the  planes  and  the  angle  which 
the  beam  makes  with  the  horizontal. 


1000' 


*  17.  A  bar  4  feet  in  length,  weighing  50  pounds,  with  its  center  of  mass  at 
the  middle,  has  one  end  against  a  smooth  vertical  wall.  The  other  end  is 
attached  to  a  rope  5  feet  in  length  which  is  fastened  to  a  point  in  the  wall. 
Find  the  position  of  equilibrium,  the  reaction  of  the  wall,  and  the  tension 
in  the  rope. 

18.  The  bar  ABt  Fig.  110,  has  its  center  of  mass  at  the  middle  of  its  length. 
The  end  B  is  against  a  smooth  wall  which  makes  an  angle  of  10  degrees 
with  the  vertical.     The  end  A  is  supported  by  a  rope.     Find  the  direction 
of  the  rope,  the  tension  in  it,  and  the  reaction  of  the  plane. 

19.  In  Fig.  Ill,  find  all  reactions  and  check.     Find  bh,  hg,  and  gf  by 
sections.     Find  all  forces  joint  at  a  time.     With  the  reactions  known,  solve 
by  Bow's  method. 


CHAPTER  VI 
COUPLES 

69.  Moment  of  a  Couple. — In  Fig.   112,  P  and  Q  are  two 

forces  in  opposite  directions  at  a  distance  a  apart.  Their 
equilibrant  is  the  force  Q  —  P  at  a  distance  x  from  Q.  The 
distance  x  is  given  by  the  equation, 

(Q  -P)x  =  Pa  .      (1) 

IfQ  =  P,Q  -  P  =  Q,x  =  infinity. 

Two  equal  and  opposite  forces  acting  on  a  body  form  a  couple. 
The  forces  of  a  couple  have  a  moment  about  any  point  and  tend 
to  produce  rotation  about  any  point.  Since  the  forces  are  equal, 
their  resultant  is  zero.  The  forces  of  a  couple  do  not  produce 
translation. 

If  the  magnitude  of  each  force  is  P  and  the  distance  between 
the  forces  is  a,  the  moment  is  Pa.     The  moment  is  the  same  with 

f  ' 

£i--4-   --1       o         f 
=* 

§TP  Q-P  TP 

FIG.  112  FIG.  113. 

respect  to  any  point  in  the  plane  of  the  forces.  This  statement 
may  be  proved  by  Fig.  113.  The  point  0,  at  a  distance  x  from 
the  downward  force  of  the  couple,  may  be  taken  as  the  origin 
of  moments.  The  sum  of  the  moments  of  the  two  forces  about 
this  origin  is 

M  =  -  Px  +  P(x  +  a)  =  Pa  (2) 

The  moment  of  the  couple  is  the  same,  no  matter  what  point  is 
taken  as  the  origin.  The  moment  of  a  couple  is  the  product  of 
either  force  multiplied  by  the  distance  between  the  lines  of  action  of 
the  two  forces. 

70.  Equivalent  Couples. — Two  forces  are  equivalent  if  either 
force  may  be  balanced  by  a  third  force  so  that  no  translation  is 
produced  in  the  body  upon  which  the  forces  act.     Two  couples 

111 


112  MECHANICS  IABT.  70 

are  equivalent  if  either  couple  may  be  balanced  by  a  third  couple 
so  that  no  rotation  is  produced  in  the  body  upon  which  the  couples 
act.  These  definitions  may  be  stated  briefly:  Forces  which  are 
balanced  by  the  same  force  are  equivalent.  Couples  which  are 
balanced  by  the  same  couple  are  equivalent^ 

In  order  that  two  forces  may  produce  equilibrium,  the  forces 
must  be  equal,  opposite,  and  along  the  same  line.  In  order  that 
two  couples  may  be  in  equilibrium,  the  magnitude  of  their  mo- 
ments must  be  equal,  the  direction  of  rotation  must  be  opposite, 
and  the  couples  must  lie  in  the  same  plane  or  in  parallel  planes. 

It  will  now  be  proved  that  a  given  couple  may  be  balanced  by 
a  second  couple  in  the  same  plane,  provided  the  moments  are 
equal  and  opposite.  If  the  moments  are  equal  and  opposite, 
the  forces  may  have  any  magnitude  and  may  be  placed  at  any 
position  in  the  plane.  A  couple  may  be  made  of  an  upward 


<?\        >R^f      \ 
P,  V-V^     P- 

1        *-"  & 


\ 


%  \         Vector  Diagram 

FIG.   114. 

force  of  6  pounds  and  a  downward  force  of  6  pounds  at  a  distance 
of  4  feet  to  the  right  of  the  first  force.  The  moment  of  this 
couple  is  24  foot-pounds  clockwise.  Any  counter-clockwise 
couple  of  24  foot-pounds  in  the  same  plane  will  produce  equilib- 
rium. This  second  couple  may  be  made  of  two  horizontal 
forces  of  24  pounds  at  a  distance  of  1  foot  apart,  with  the  upper 
force  toward  the  left,  or  it  may  be  made  of  two  forces  of  12 
pounds  at  a  distance  of  2  feet  apart,  or  of  two  forces  of  8  pounds 
at  a  distance  of  3  feet,  or  of  any  other  combination  whose  moment 
is  24  foot-pounds  counter-clockwise. 

In  Fig.  114,  PI  and  P2  are  equal  forces  of  magnitude  P  at  a 
distance  a  apart.  These  forces  form  a  clockwise  couple  of  magni- 
tude Pa.  A  second  pair  of  forces,  Qi  and  Q2,  each  of  magnitude 
Q,  at  a  distance  b  apart,  form  a  counter-clockwise  couple  of 
moment  Qb.  It  will  be  proved  that  these  couples  produce 
equilibrium  if  the  magnitude  of  the  moment  Pa  is  equal  to  the 
magnitude  of  the  moment  Qb, 


CHAP.   VI]  COUPLES  113 

Extend  the  lines  of  the  forces  until  the  line  of  the  force  PI 
intersects  the  line  of  the  force  Qi,  and  the  line  of  the  force  P2 
intersects  the  line  of  the  force  Q2.  The  resultant  of  the  forces 
Pi  and  Qi  at  A  is  a  force  RI.  The  direction  and  magnitude  of  the 
force  RI  are  given  by  the  vector  diagram,  E,  F,  G,  of  Fig.  114,  II. 
The  resultant  of  P2  and  Q2  at  C  is  a  force  R2.  The  magnitude 
and  direction  of  R2  are  given  by  the  lower  force  triangle  of  Fig. 
114,  II.  Since  PI  is  equal  and  opposite  to  P2,  and  Qi  is  equal  and 
opposite  to  Q2,  it  is  evident  from  Fig.  114,  II,  that  RI  is  equal  and 
opposite  to  R2.  If  the  resultant  RI  through  A  of  the  space  dia- 
gram falls  on  the  same  line  as  the  resultant  R2  through  C,  the 
forces  will  balance  and  the  couples  will  be  in  equilibrium.  This 
condition  is  satisfied  if  the  line  of  RI  passes  through  the  point  C. 

In  the  space  triangle  ABC,  and  the  force  triangle  EFG,  the 
angle  at  B  is  equal  to  the  angle  at  F.  If  this  angle  is  represented 
by  0; 

AB  sin  6  =  b,  (1) 

BC  sin  0  =  o,  (2) 

AB  -  b  m 

"  a 


The  triangles  ABC  and  EFG  are  similar  if 

AB       EF 
BC=FG' 

and  the  resultant  RI  on  the  space  diagram  falls  on  the  line  AC. 
Substituting  from  Equation  (3) 

6      EF      P  (5) 


a       FG      Q 

Pa  =  Qt  (6) 

Equation  (6)  states  that  the  moments  of  the  two  couples  are 
equal,  when  ABC  and  EFG  are  similar. 

Since  the  couple  Pa  may  be  balanced  by  any  other  couple  of 
equal  moment  and  opposite  direction  in  the  same  plane,  it 
follows  that  any.  two  couples  in  the  same  plane  are  equivalent  if 
moments  are  equal  in  magnitude  and  sign. 

If  the  moment  Pa  of  Fig.  114  does  not  equal  the  moment  Qb, 
the  resultant  RI  will  still  be  equal  and  opposite  to  the  resultant 
R2.  The  two  resultants  will  not,  however,  lie  on  the  same  line 
but  will  form  a  new  couple  of  moment  Pa  —  Qb. 

8 


114  MECHANICS  '  [ART.  71 

Problems 

1.  A  couple  is  made  up  of  a  horizontal  force  of  12  pounds  toward  the 
right  and  an  equal  force  toward  the  left  at  a  distance  of  3  inches  below  the 
first  force.     A  second  couple  is  made  up  of  a  force  of  8  pounds  upward  at 
an  angle  of  30  degrees  to  the  right  of  the  vertical  and  an  equal  force  in  the 
opposite  direction  at  a  distance  of  5  inches,  so  placed  that  the  moment  is 
counter-clockwise.     Draw  a  space  diagram  similar  to  Fig.  114  to  the  scale 
of  1  inch  =  1  inch.     Construct  the  force  diagram  similar  to  EFG  of  Fig.  114 
to  the  scale  of  1  inch  =  4  pounds,  and  find  RI.     Through  A  and  C  draw  lines 
parallel  to  RI.     Measure  the  distance  between  these  lines  and  multiply  by 
RI.     Compare  the  product  with  the  difference  of  the  two  moments. 

2.  Solve  Problem  1  if  the  forces,  of  the  second  couple  are  so  placed  that 
its  moment  is  clockwise. 

3.  A  clockwise  couple  is  made  up  of  two  horizontal  forces  of  10  pounds 
each  at  a  distance  of  6  inches  from  each  other.     A  counter-clockwise  couple 
in  the  same  plane  is  made  up  of  two  forces  of  15  pounds  each  at  a  distance  of 
4  inches  apart.     Find  their  resultant  graphically. 

It  will  be  shown  in  Art.  106  that  couples  in  parallel  planes 
which  are  equal  in  magnitude  and  have  the  same  signs  are  equiva- 
lent. 

71.  Algebraic  Addition  of  Couples. — Couples  in  the  same  plane 
may  be  added  graphically  by  methods  of  Art.  70.  It  will 


II 

FIG.  115. 

now  be  proved  that  the  combined  moment  of  two  couples  in 
the  same  plane  is  the  algebraic  sum  of  the  moments  of  the  couples. 
If  Pa  is  one  couple  and  Qb  is  another  couple  in  the  same  plane, 
the  resultant  moment  is  Pa  +  Qb  when  the  moments  are  in  the 
same  direction,  and  the  resultant  moment  is  Pa  —  Qb  when  the 
moments  are  in  opposite  directions. 

There  are  two  couples  in  Fig.  115.     If  —  =  R,  the  couple 

Qb  may  be  replaced  by  a  second  couple  Ra.  The  forces  of  the 
second  couple  are  RI  and  R2,  each  of  magnitude  R.  According 
to  Art.  70,  this  couple  Ra  may  be  placed  anywhere  in  the  plane  of 
the  original  couples.  In  Fig.  115,  II,  the  force  RI  is  placed  in  the 
line  of  the  force  PI,  and  the  force  R2  is  placed  in  the  line  of  the 


CHAP.   VI] 


COUPLES 


115 


force  P2.  The  resultant  force  in  each  of  these  lines  is  now  P  +  R 
and 

M  =  (P  +  R)a  =  Pa  +  Ra  =  Pa  +  Qb 

This  equation  proves  the  proposition.  Any  number  of  couples 
may  be  added  in  the  same  way. 

Since  the  moment  of  a  couple  is  the  same  about  any  point  in 
its  plane,  it  might  be  regarded  as  self-evident  that  the  combined 
moment  of  several  couples  is  the  algebraic  sum  of  the  separate 
moments.  The  proof  here  given  is,  however,  more  satisfactory 
to  most  readers,  than  this  brief  statement. 

72.  Equilibrium  by  Couples. — In  many  problems  of  equilib- 
rium, the  forces  may  be  grouped  to  form  two  equal  and  opposite 


FIG.  116. 

couples  in  the  same  plane.  Figure  116  represents  an  arrangement 
for  demonstrating  the  equilibrium  of  couples.  A  rigid  body 
is  made  up  of  a  bar  and  a  wheel  fastened  together.  The  bar 
is  supported  by  a  small  cylinder  which  rolls  on  a  plane  surface. 
The  reaction  of  the  small  cylinder  is  vertical.  This  reaction 
together  with  the  weight  of  the  system  forms  one  couple.  Two 
cords  are  passed  partly  around  the  wheel  and  fastened  to  it. 
One  cord  runs  horizontally  toward  the  left  and  is  attached  to  a 
post.  The  second  cord  runs  horizontally  toward  the  right, 
passes  over  a  smooth  pulley,  and  supports  a  mass  of  P  pounds. 
P  =  T  and  R  =  W  For  equilibrium 

Pa  =  Wb 
The  plane  supporting  the  cylinder  may  be  a  platform  scale. 


116  MECHANICS  [ART.  73 

If  the  scale  poise  is  set  to  equal  the  load  W  and  the  weight  of 
the  small  cylinder,  the  beam  will  be  in  balance  when  the  cords 
are.  parallel. 

Figure  117  shows  a  beam  with  loads  P  and  Q  near  the  ends. 


k- 
\ 


9 

—  ><  ------  b  -----  >\ 


L 


FIG.  117. 


Neglecting  the  weight  of  the  beam,  the  .  reaction  S  =  P  +  Q. 
The  load  P  at  the  left  end  together  with  an  equal  amount  of  the 
reaction  at  the  middle  forms  a  couple  of  moment  Pa.  The  load 
Q  at  the  right  end,  together  with  the  remainder  of  the  reaction, 


forms  an  opposite  couple  of  moment  Qb.     For  equilibrium  these 
couples  are  equal. 

In  Fig.  118,  the  two  forces  P  and  Q  are  not  parallel.  The 
equilibrant  S  may  be  resolved  into  two  components,  which 
are  equal  and  opposite  to  P  and  Q,  respectively.  The  beam 
is  then  subjected  to  two  couples  of  moments  Pa  and  Qb. 

Problems 

1.  What  are  the  two  equal  couples  in  Fig.  96  of  Art.  63? 

2.  What  are  the  two  equal  couples  of  Fig.  88? 

3.  In  Fig.  89,  when  AC  and  BD  are  parallel,  what  are  the  two  equivalent 
couples? 

73.  Reduction  of  a  Force  and  a  Couple  to  a  Single  Force. — In 
Fig.  119,  I,  there  is  a  single  force  P,  and  a  counter-clockwise 
couple  in  the  same  plane  made  up  of  two  equal  forces  Qi  and 


CHAP.   VI]  COUPLES  117 

Q2  at  a  distance  b  apart.  The  moment  of  this  couple  is  Qb. 
If  Pa  =  Qb,  this  couple  may  be  replaced  by  a  couple  made  up  of 
two  equal  forces  PI  and  P2  at  a  distance  a  apart.  By  Art.  70 
these  forces  may  be  placed  anywhere  in  the  plane  of  the  couple. 
The  force  PI  may  be  placed  in  the  line  of  action  of  the  single  force 
P,  and  the  force  P2  at  a  distance  a  from  that 
line  on  the  side  which  makes  the  moment  ^2A 

counter-clockwise.     The    force   PI,    Fig.    119, 
II,  balances  the  force  P.     The  remaining  force     p  /  j    ^Q, 
P2,  at  a  distance  a  from  the  position  of  the  force      / 
P,  replaces  the  force  and  the  couple.  p£       r 

A  force  and  a  couple  in  the  same  plane  are  /       ypz 

equivalent  to  a  single  force.     The  direction  and         4        / 
magnitude  of  the  single  force  are  the  same  as       /        / 
those  of  the  line  of  action  of  the  original  force. 
Its  distance  from  the  line  of  action  of  the  original 
force  is  such  that  its  moment  about  any  point  in  that  line  is  equal  in 
magnitude  and  sign  to  the  moment  of  the  original  couple. 

Example  ^§ 

A  vertical  force  of  20  pounds  upward  is  applied  at  the  point  x  =  4  ft. 
A  horizontal  force  of  5  pounds  toward  the  right  is  applied  at  y  =  2  ft.  and 
an  equal  horizontal  force  toward  the  left  is  applied  at  y  =  12  ft.  Find  the 
location  of  the  single  force  which  is  equivalent  to  this  force  and  couple. 

The  moment  of  the  couple  is  50  foot-pounds. 

|  '  >  |!  =  2.5ft. 

The  single  force  is  20  pounds  upward  at  a  distance  of  2.5  feet  from  the 
vertical  line  through  the  point  x  =  4  ft.  Since  the  couple  is  counter- 
clockwise, the  distance  of  2.5  feet  must  be  measured  toward  the  right  from 
the  line  x  =  4  ft.  The  resultant  force  lies  in  the  line  x  =  6.5  ft. 

Problems 

1.  A  force  of  12  pounds,  along  the  line  x  =  2  ft.,  is  combined  with  a 
couple  made  up  of  a  horizontal  force  of  8  pounds  toward  the  right,  along 
the  line  y  =  1  ft.,  and  an  equal  and  opposite  force,  along  the  line  y  =  13  ft. 
Find  the  location  of  the  single  force  which  is  equivalent  to  the  force  and 
the  couple. 

2.  Solve  Problem  1  graphically  to  the  scale  of  1  inch  =  4  feet,  and  1 
inch  =  8  pounds.     Through  the  intersection  of  the  line  of  action  of  the 
vertical  force  with  the  line  of  action  of  one  of  the  horizontal  forces  draw  a 
line  parallel  to  the  direction  of  their  resultant.     Through  the  intersection 
of  this  line  with  the  line  of  the  third  force  on  the  space  diagram,  draw  a 
line  parallel  to  the  direction  of  the  resultant  of  all  three  forces. 


118  MECHANICS  [ART.  74 

3.  Find  the  single  force  which  can  be  substituted  for  a  horizontal  force  of 
16  pounds  directed  toward  the  right  and  a  clockwise  couple  of  48  foot-pounds. 

Ans.      16  pounds  towards  the  right  through  a  point  3  ft.  above  the  original 
force. 

4.  An  upward  vertical  force  of  8  pounds  acts  along  the  F-axis,  and  a 
second  upward  vertical  force  of  4  pounds  acts  along^  the  line  x  =  6  ft.     A 
couple  of  60  foot-pounds  clockwise  and  a  couple  of  &  foot-pounds  counter- 
clockwise act  on  the  body  in  the  same  plane  with  these  forces.     Find  the 
single  force  which  will  replace  all  of  these. 


H 

A 


74.  Resolution  of  a  Force  into  a  Force  and  a  Couple. — Figure 
120  shows  a  single  force  P.  It  is  desired  to  replace  this  force  by 
the  force  and  a  couple  of  moment  Pa.  In  Fig.  120, 
C  is  a  point  at  a  distance  a  from  the  line  of  action 
of  the  force  P.  At  C  are  applied  two  opposite  forces, 
Pi  and  P2,  each  of  which  is  equal  in  magnitude  to 
the  force  P,  and  along  a  parallel  line.  Since  these 
forces  balance  each  other,  they  have  no  effect  upon 
the  equilibrium  of  the  body  upon  which  they  act. 
The  force  P|  and  the  force  P  form  a  couple  of  moment  Pa. 
The  force  Pj^  which  is  equal  and  parallel  to  the  original  force 
P,  stands  alone  as  the  single  force  required. 

A  single  force  may  be  replaced  by  an  equal  force  in  the  same  direc- 
tion through  any  point  in  its  plane,  and  a  couple,  the  moment  of 
which  is  the  same  in  magnitude  and  direction  as  the  moment  of 
the  original  force  about  that  point. 

Problems 

1.  A  force  of  12  pounds  upward  acts  at  the  point  x  —  3.     Replace  this 
force  by  an  equal  force  through  the  origin  and  a  couple. 

Ans.  12  pounds  upward  through  the  origin  and  a  counter-clockwise 
couple  of  36  units. 

2.  A  force  of  16  pounds  at  an  angle  of  45  degrees  to  the  right  of  the 
vertical  upward  acts  at  the  point  x=  3  ft.,  y  =  5  ft.     Replace  by  a  force 
through  the  origin  and  a  couple. 

Ans.  16  pounds  through  the  origin  at  an  angle  of  45  degrees  to  the  right 
of  the  vertical,  and  a  counter-clockwise  couple  of  22.62  foot-pounds. 

3.  A  force  of  &Q  pounds  at  an  angle  of  25  degrees  to  the  right  of  the 
vertical  is  applied  at  the  point  x  =  2  ft.,  y  =  3  ft.  and  a  force  of  15  pounds 
at  an  angle  of  60  degrees  to  the  right  of  the  vertical  is  applied  at  the  point 
x  =  2  ft.,  y  =  8  ft.     Replace  these  by  a  single  force  at  the  origin  and  a 
single  couple. 

Replace  each  force  by  a  single  force  at  the  origin  and  a  single  couple. 
Add  the  two  couples  and  find  the  resultant  of  the  two  forces. 


CHAP.    VI]  COUPLES  119 

The  principles  of  this  article  afford  a  method  of  finding  the 
resultant  of  a  set  of  non-concurrent,  coplanar  forces.  Let 
Pi,  P2,  PS,  etc  be  a  set  of  forces  in  the  same  plane.  Each  force 
may  be  replaced  by  an  equal  force  in  the  same  direction  applied 
at  some  convenient  point  and  a  couple.  Since  all  the  forces  are 
now  applied  at  the  same  point,  they  may  be  treated  as  concurrent 
and  their  resultant  found  by  means  of  the  force  polygon  or  cal- 
culated by  the  methods  of  Art.  42.  The  couples  may  be  added 
algebraically.  Their  sum  is  a  single  couple. 

The  resultant  of  a  set  of  non-concurrent  forces  in  the  same 
plane  is  equivalent  to  a  single  force  and  a  single  couple.  Since 
a  force  and  a  couple  in  the  same  plane  may  be  reduced  to  a  single 
force  by  the  methods  of  Art.  73,  it  follows  that,  in  general,  the 
resultant  of  a  set  of  non-concurrent,  coplanar  forces  is  a  single 
force.  It  sometimes  happens  that  the  resultant  force  is  zero, 
while  the  resultant  couple  is  not  zero.  This  resultant  couple 
can  hot  be  reduced  to  a  single  force. 

A    set    of    non-concurrent,    coplanar  forces  may  always  be 
reduced  to  a  single  force  or  to  a  single  couple. 
'     While  the  steps  of  the  proof  are  different,  the  results  of  this 
article  are  practically  the  same  as  the  method  of  finding  the 
resultant,  which  is  given  in  Art.  60. 

75.  Summary. — Two  equal,  opposite  forces  form  a  couple. 
The  moment  of  a  couple  is  the  product  of  either  force  multiplied 
by  the  distance  between  them.  The  moment  of  a  couple  is  the 
same  with  respect  to  any  origin. 

Two  couples  in  the  same  plane  are  equivalent  if  their  moments 
are  equal  in  magnitude  and  sign. 

The  combined  moment  of  several  couples  in  the  same  plane  is 
the  algebraic  sum  of  the  separate  moments. 

A  force  and  a  couple  in  the  same  plane  may  be  replaced  by  a 
single  force  which  is  equal  in  magnitude  and  direction  to  the 
original  force,  and  is  so  located  that  its  moment  about  any  point 
in  the  line  of  the  original  force  is  equal  to  the  moment  of  the  couple. 

A  single  force  may  be  replaced  by  an  equal  force  in  the  same 
direction  through  any  point  in  its  plane,  and  a  couple  whose 
moment  is  equivalent  to  the  moment  of  the  force  about  that  point. 

A  set  of  non-concurrent,  coplanar  forces  may  be  replaced  by  a 
single  force  and  a  single  couple.  These  may  be  reduced  to  a 
single  force,  except  when  the  resultant  force  is  zero.  In  this 
case,  the  final  resultant  is  a  couple. 


CHAPTER  VII 
GRAPHICS  OF  NON-CONCURRENT  FORCES 

76.  Resultant  of  Parallel  Forces.— In  Art.  55,  a  method  was 
given  for  finding  the  resultant  of  parallel  forces  in  the  same  direc- 
tion. In  Fig.  76,  the  rigid  bar  was  assumed  to  be  connected  to 
two  ropes  or  hinged  rods,  and  the  equilibrant  was  found  as  a 
problem  of  connected  bodies  with  concurrent  forces.  This  proof 
was  given  because  it  is  concrete  and  because  it  fits  the  methods 
of  the  preceding  chapter.  A  more  abstract  proof  is  generally 
used.  This  will  now  be  given. 

Figure  121,  I,  shows  two  parallel  forces  P  and  Q  which  are 
applied  to  the  ends  of  a  rigid  bar.  The  force  P  at  the  left  end  is 


Force  Polygon 


Fio.   121. 


resolved  into  two  components  RI  and  S,  which  are  not  at  right 
angles  to  each  other.  These  are  shown  in  the  triangle  at  the 
left  of  the  bar.'  The  component  S  is  parallel  to  the  bar.  The 
force  Q  at  the  right  end  is  resolved  into  two  components  —  S  and 
R2.  The  component  —  S  is  equal  and  opposite  to  thec  omponent 
S  at  the  left  end.  On  the  space  diagram,  it  is  seen  that  the  force 
S  balances  the  force  —  S  and  the  components  RI  and  R2  remain  to 
act  on  the  bar.  The  resultant  of  RI  and  R2  must  pass  through 
the  intersection  of  the  broken  lines  of  the  space  diagram,  which 
are  drawn  parallel  to  the  corresponding  forces  of  the  force  tri- 

120 


CHAP.  VII]         NON-CONCURRENT-FORCE  121 

angles.  Figure  121,  II  shows  the  two  force  triangles  combined, 
as  they  are  generally  drawn.  The  components  S  and  —  S  fall  on 
the  same  line.  The  resultant  of  RI  and  R2  is  the  force  P  +  Q, 
in  the  direction  of  the  original  forces. 

Figure  121,  III,  shows  the  usual  method  of  construction  for  the 
position  of  the  resultant.  The  parallel  forces  are  P  and  Q.  It 
is, not  necessary  to  draw  the  body  upon  which  they  act,  or  to 
consider  the  direction  of  the  line  which  joins  their  point  of  appli- 
cation. The  force  diagram  is  first  drawn.  The  distances  ab  = 
P  and  be  =  Q  are  laid  off  on  a  vertical  line.  Then  a  point  0  is 
selected  to  one  side  of  this  line,  and  connected  to  the  points  a,  b, 
and  c  by  lines  ao,  bo,  and  co.  The  figure  thus  obtained  is  the 
same  as  the  force  diagram  of  Fig.  121,  II.  A  point  No.  1  is 
chosen  on  the  line  of  the  force  P  and  a  line  is  drawn  through  it 
parallel  to  ao  and  a  second  line  parallel  to  bo.  The  line  parallel 
to  bo  is  extended  till  it  intersects  the  line  of  the  force  Q  at  the 
point  No.  2.  Through  point  No.  2,  a  line  is  drawn  parallel  to 
the  line  co  of  the  force  diagram.  The  intersection  -of  the  line 
parallel  to  ao  with  the  line  parallel  to  co  gives  a  point  on  the  line 
of  the  resultant.  The  resultant  is  equal  and  parallel  to  ac  of 
the  force  diagram. 

The  space  diagram  of  Fig.  121,  III  is  called  a  string  polygon,  or 
funicular  polygon.  The  solid  lines  of  the  drawing  give  the 
directions  which  three  connected  strings  could  take  when  sup- 
porting the  loads  P  and  Q  at  a  definite  distance  apart.  The 
diagram  is  lettered  by  Bow's  method  to  correspond  with  the  force 
diagram  of  the  figure.  (The  line  P  +  Q  does  not  represent  a 
force,  but  only  the  position  of  the  resultant.)  The  arrows  show 
the  directions  required  for  equilibrium. 

The  point  0  of  the  force  polygon  of  Fig.  121,  III,  is  called  the 
pole.  The  lines  ao,  bo,  and  co  are  rays.  The  position  of  the  pole 
may  be  chosen  at  any  point.  It  should  be  so  selected  that  lines 
parallel  to  the  rays  will  intersect  at  convenient  positions  on  the 
string  polygon. 

Problems 

1.  Two  vertical  loads  of  15  pounds  and  20  pounds  are  6  inches  apart. 
Draw  a  force  polygon  to  the  scale  of  1  inch  =  5  pounds  and  put  the  pole  5 
inches  to  the  right  of  the  load  line.     Draw  the  string  polygon  to  the  scale  of 
1  inch  =  1  inch  and  find  the  distance  of  the  resultant  from  the  loads. 
Check  by  moments.     Change  the  position  of  the  pole  and  solve  again. 

2.  Solve  Problem  1  with  the  pole  to  the  left  of  the  load  line.     Explain 
the  result. 


122 


MECHANICS 


[ART.   77 


3.  Draw  the  force  polygon  and  the  funicular  polygon  for  three  vertical 
forces:  13  pounds  downward  at  0  feet,  10  pounds  downward  at  4  feet,  and 
18  pounds  downward  at  12  feet.     Use  scale  of  1  inch  =  5  pounds  and  1 
inch  =  2  feet.     Compare  with  Fig.  122  which  is  drawn  for  a  different  load- 
ing.    Locate  the  resultant  force  and  check  by  moments. 

4.  Three  weights  of  10  pounds  at  0  feet,  15  pounds  at  4  feet,  and  10 
pounds  at  8  feet,  are  connected  by  cords  and  supported  by  two  cords.     The 
cords  supporting  the  first  and  last  weights  make  angles  of  45  degrees  with 
the  horizontal.     Find  the  direction  and  length  of  the  cords  connecting  the 
weights. 

5.  Given  three  vertical  loads:  12  pounds  at  0  feet,  16  pounds  at  5  feet, 
and  20  pounds  at  8  feet.     Find  the  location  of  the  resultant  graphically  and 
check  by  moments.     If  the  cord  supporting  the  12  pounds  makes  an  angle 
of  60  degrees  to  the  left  of  the  vertical,  and  the  cord  supporting  the  20 
pounds  maks  an  angle  50  degrees  to  the  right  of  the  vertical,  what  is  the 
direction  and  length  of  the  connecting  cords? 


Line  of  \ 

Resultant-^  \ 

A       7^ "      I  I 


String  Polygon 


Force 
Polygon 


FIG.  122. 


77.  Resultant  of  Non-parallel  Forces.— Art.  59  gives  a  method 
of  finding  the  resultant  of  non-parallel  forces  graphically.  It 
frequently  happens  that  some  of  the  forces  are  so  nearly  parallel 
that  this  method  requires  an  extremely  large  space  diagram  to 
get  the  intersections.  In  such  a  case  the  results  are  likely  to  be 
inaccurate.  In  most  cases  of  non-parallel  forces,  especially 
when  there  are  more  than  three  forces,  it  is  advisable  to  use  the 
method  of  Art.  76.  This  differs  from  the  method  of  Art.  59 
only  in  the  fact  that  the  first  and  last  forces  are  resolved  into 
components,  with  one  component  of  the  first  force  equal  and 
opposite  to  one  component  of  the  last  force,  and  acting  along 
the  same  line.  By  properly  choosing  the  location  of  the  pole 
in  the  force  diagram,  the  rays  may  be  given  such  directions  that 
the  forces  which  they  represent  will  make  large  angles  with  the 
directions  of  the  applied  loads  on  the  space  diagram.  The 
intersections  will  then  be  found  accurately. 


CHAP.   VII] 


NON-CONCURRENT  FORCES 


123 


Example 

A  rectangular  board  in  a  vertical  plane  is  4  feet  wide  horizontally  and 
3  feet  high.  The  board  weighs  20  pounds  and  its  center  of  mass  is  at  the 
center.  A  cord,  30  degrees  to  the  left  of  the  vertical  downward  is  applied 
at  the  lower  left  corner  and  exerts  a  pull  of  12  pounds.  A  cord  at  the  lower 
right  corner  makes  an  angle  of  15  degrees  to  the  right  of  the  vertical  down- 
ward and  exerts  a  pull  of  15  pounds.  A  cord  at  45  degrees  to  the  right  of 
the  vertical  downward  is  attached  to  the  upper  right  corner  and  exerts  a 
pull  16  pounds.  The  board  is  held  in  equilibrium  by  a  single  cord.  Find 
the  direction  and  location  of  this  cord  and  the  force  which  it  exerts. 

Figure  123  gives  the  solution.  The  original  drawing  was  made 
to  the  scale  of  1  inch  =  1  foot  on  the  space  diagram  and  1  inch  = 
10  pounds  on  the  force  diagram.  The  space  diagram,  showing 
the  position  and  direction  of  the  known  forces,  was  first  drawn. 
Next,  the  load  line  abode  of  the  force  polygon  was  laid  off.  The 
pole  was  taken  4  inches  from  the  middle  of  the  vertical  line  be. 


20 


FIG.   123. 


This  position  makes  the  ray  ao  nearly  normal  to  the  direction 
of  the  force  ab,  and  the  ray  eo  nearly  normal  to  the  force  de,  and 
makes  large  angles  between  the  rays  and  the  forces  which  they 
intersect.  Next,  a  point  No.  1  was  chosen  in  the  line  of  the  12- 
pound  force  on  the  space  diagram.  Through  point  No.  1,  a 
line  was  drawn  toward  the  left  parallel  to  the  ray  ao,  and  a  second 
line  toward  the  right  parallel  to  the  ray  bo.  This  second  line 
intersects  the  line  of  the  20-pound  force  at  point  No.  2.  This 
operation  was  continued  for  all  the  rays.  Finally,  the  first  line 
through  point  No.  1  was  extended  to  the  right  and  the  last  line 
through  point  No.  4  was  extended  to  the  left.  The  intersection 


124  MECHANICS  [ART.  78 

of  these  lines  is  point  No.  5,  which  lies  on  the  line  of  the  resultant 
(or  the  equilibrant)  of  all  these  forces.  The  resultant  of  all 
these  forces  is  the  line  ae  of  the  force  polygon.  The  broken  line 
through  point  No.  5,  drawn  parallel  to  ae,  gives  the  line  of  action 
of  the  resultant.  A  single  cord  attached  to  the  board  at  any 
point  on  this  line  may  hold  it  in  equilibrium. 

Problems 

1.  Solve  the  Example  above  graphically,  choosing  the  pole  3  inches  from 
c  on  the  horizontal  line  through  c. 

2.  The  following  forces  act  on  a  rigid  body:  20  pounds  18  degrees  to  the 
left  of  the  vertical  downward,  applied  at  the  point  (—4,  2);  16  pounds 
vertically  downward,  applied  at  the  point  (  —  1,  0);  18  pounds  25  degrees 
to  the  right  of  the  vertical  downward,  applied  at  the  point  (2,  1);  and  24 
pounds  40  degrees  to  the  right  of  the  vertical  downward,  applied  at  the 
point  (4,  0).     Find  the  magnitude,  direction,  and  location  of  the  resultant 
graphically. 

3.  Solve  the  Example  of  Art.  60  by  the  graphical  method. 

78.  Parallel  Reactions. — When  a  body  is  subjected  to  a  set  of 
forces  in  the  same  plane,  it  may  be  held  in  equilibrium  by  a 
single  force  in  that  plane,  except  when  the  resultant  is  a  couple. 
This  equilibrant  is  equal  and  opposite  to  the  resultant  of  all  the 
forces  and  lies  in  the  same  line.  The  line  of  the  resultant  in 
the  space  diagrams  of  the  problems  of  Art.  77  gives  the  location 
of  a  series  of  points  at  which  the  body  may  be  supported  by 
a  single  smooth  hinge. 

Usually  such  bodies  are  supported  at  two  points.  When  the 
loads  are  all  parallel,  the  reactions  at  these  points  are  generally 
parallel.  In  most  cases  the  positions  of  the  supports  are  given 
and  the  problem  is  that  of  finding  the  magnitude  of  the  reactions. 

The  resultant  of  all  the  parallel  loads  may  be  found  by  the 
methods  of  Art.  76.  This  resultant. is  equal  and  opposite  to 
the  equilibrant,  which,  in  turn,  is  the  resultant  of  the  reactions. 
The  problem  of  finding  these  reactions  resolves  itself  into  the 
problem  of  finding  two  parallel  forces  which  have  a  given  resul- 
tant and  which  pass  through  known  points.  It  is  the  converse 
of  the  problem  of  finding  the  resultant  of  two  known  forces. 

Figure  124  gives  the  method  of  finding  the  resultant  of  two 
reactions,  RI  and  R2.  Suppose  that  the  magnitude  and  position 
of  the  resultant  S  are  known;  that  the  positions  of  the  reactions 
RI  and  R2  are  also  known;  and  that  it  is  desired  to  find  the 


CHAP.  VII]         NON-CONCURRENT  FORCES 


125 


magnitude  of  these  reactions.  First,  draw  the  space  diagram, 
giving  the  positions  of  S,  RI,  and  R2.  Then,  begin  the  force 
polygon  by  laying  off  the  line  ca  of  length  and  direction  equal  to 
the  resultant.  Choose  a  point  as  the  pole  and  draw  the  rays  ao 
and  co.  Next,  choose  a  point  No.  1  on  the  line  of  action  of  RI 
and  draw  a  line  parallel  to  the  ray  ao  intersecting  the  line  of 
the  resultant  at  point  No.  3.  Through  this  point  draw  a  line 
parallel  to  the  ray  co  and  locate  its  point  of  intersection  with  the 
line  of  R£.  This  is  point  No.  2.  Finally,  join  point  No.  1  with 
point  No.  2,  and  draw  a  line  through  the  pole  parallel  to  the  line 


Force 
Polygon 


String  Polygon 

FIG.  124. 


1-2.  This  line  intersects  the  line  ac  at  the  point  b.  The  length 
ab  on  the  force  polygon  gives  the  left  reaction,  and  the  length  be 
gives  the  right  reaction. 

Problems 

1.  A  horizontal  beam  12  feet  long,  weighing  60  pounds,  with  its  center 
of  mass  5  feet  from  the  left  end,  is  supported  at  the  ends.     Find  the  reactions 
of  the  supports  graphically.     Check  by  moments. 

2.  A  beam  20  feet  long,  weighing  50  pounds,  has  its  center  of  mass  8 
feet  from  the  left  end.     It  is  supported  by  means  of  vertical  ropes  at  the 
ends.     Find  the  tension  in  each  rope. 

3.  A  horizontal  beam  12  feet  long,  weighing  40  pounds,  with  its  center  of 
mass  at  the  middle,  is  supported  by  vertical  ropes  at  the  ends.     It  carries  a 
load  of  30  pounds  3  feet  from  the  left  end,  and  a  load  of  50  pounds  2  feet 
from  the  right  end.     Find  the  reactions. 

The  resultant  force  may  be  found  by  the  methods  of  Art. 
76;  and  the  reactions,  as  in  the  Example  above.  The  two  proc- 
esses are  generally  combined.  Figure  125  shows  the  method.  On 
the  space  diagram  draw  the  lines  of  the  loads  and  the  reactions. 
Then  lay  off  the  loads  on  the  force  polygon  beginning  with  the 
load  at  the  left  of  the  space  diagram.  Choose  a  pole  and  draw  the 


126 


MECHANICS 


[ART.   78 


rays.  Beginning  with  the  left  load  of  30  pounds  select  point  No.  1 . 
Draw  a  line  through  point  No.  1  parallel  to  bo.  This  line  inter- 
sects the  line  of  the  40-pound  load  at  point  No.  2.  Through 
point  No.  2  draw  the  next  line  parallel  to  the  ray  co,  intersecting 
the  line  of  the  50  pound  load  at  point  No.  3.  Through  point 
No.  1  draw  a  line  parallel  to  ao  and  through  point  No.  3  draw  a 
line  parallel  to  do.  If  these  lines  were  carried  downward,  their 
intersection  would  give  a  point  on  the  line  of  the  resultant. 
All  this  is  exactly  the  same  as  the  method  of  Art.  76. 


Space  Diagram      Funicular  Polygon 

FIG.   125. 


Force 
Polygon 


It  is  not  necessary,  however,  to  find  the  location  of  the  resul- 
tant. The  line  ad  of  the  force  polygon  gives  its  magnitude  and 
direction.  The  rays  ao  and  do  are  components  of  the  resultant. 
Draw  a  line  through  point  No.  1  parallel  to  the  ray  ao  and  extend 
it  till  it  intersects  the  line  of  the  left  reaction  at  point  No.  4. 
Similarly,  draw  a  line  through  point  No.  3  parallel  to  the  ray  do 
and  extend  it  till  it  intersects  the  line  of  the  right  reaction  at 
point  No.  5.  Since  these  lines  would  intersect  on  the  resultant 
if  they  were  extended  in  the  opposite  directions,  and  since  their 
directions  are  those  of  the  first  and  -last  rays,  ao  and  do,  they 
correspond  to  the  lines  of  Fig.  124.  Connect  point  No.  4  to 
point  No.  5  by  a  straight  line  and  draw  a  ray  parallel  to  this 
line.  The  intersection  of  this  ray  with  the  load  line  divides  ad 
into  two  parts  ae  and  ed.  The  lengths  of  these  parts  give  the 
reactions. 

Problems 

4.  A  beam  20  feet  long,  with  its  center  of  mass  at  the  middle,  weighs  60 
pounds.  It  is  supported  in  a  horizontal  position  by  a  vertical  rope  at  each 
end,  and  carries  a  load  of  40  pounds  4  feet  from  the  left  end,  and  a  load  of 
50  pounds  4  feet  from  the  right  end.  Find  the  reactions  graphically  and 
check  by  moments. 


CHAP.   VII] 


NON-CONCURRENT  FORCES 


127 


6.  Solve  Problem  1  if  the  second  rope  is  6  feet  from  the  right  end. 

6.  Solve  Problem  3  if  the  beam  is  supported  at  the  left  end  and  4  feet 
from  the  left  end. 

7.  The  beam  of  Problem  4  has  one  support  at  the  left  end,  where  the 
vertical  reaction  is  80  pounds.     Locate  the  second  support  graphically. 
Check  by  moments. 

79.  Non-parallel  Reactions.  When  the  loads  applied  to  a  rigid 
body  are  not  parallel,  the  reactions  usually  are  not  parallel. 
Sometimes  when  the  loads  are  parallel  the  reactions  are  not 
parallel.  Generally,  the  line  of  action  of  one  reaction,  and  the 
location  of  a  single  point  in  the  line  of  the  other  are  known.  The 
method  of  solution  is  the  same  as  for  parallel  reactions,  except 
that  the  point  of  beginning  of  the  funicular  polygon  is  fixed  by  the 
conditions  of  the  problem. 

Example 

A  beam  10  feet  long,  weighing  15  pounds,  with  its  center  of  mass  at  the 
middle,  is  hinged  at  the  left  end  and  supported  at  the  right  end  by  a  cord 
which  makes  an  angle  of  30  degrees  to  the  left  of  the  vertical.  The  right 
end  of  the  beam  is  elevated  10  degrees  above  the  horizontal.  The  beam 


10 

Space  Diagram      Funicular  Polygon 

FIG.  126. 


Force 
Polygon 


carries  a  vertical  load  of  10  pounds  2  feet  from  the  left  end,  and  a  load  of 
20  pounds  at  an  angle  of  1*5  degrees  to  the  right  of  the  vertical  downward 
at  a  point  2  feet  from  the  right  end.  Solve  for  the  reactions  graphically, 
using  the  scale  of  1  inch  =  2  feet  on  the  space  diagram,  and  1  inch  =  10 
pounds  on  the  force  diagram. 

First,  construct  the  load  line  and  the  rays  as  in  the  preceding 
problems.  In  Fig.  125,  the  line  of  action  of  the  left  reaction  was 
known  and  the  point  No.  4  could  fall  anywhere  on  that  line. 
In  Fig.  126,  the  left  reaction  passes  through  the  hinge,  but  its 
direction  is  not  known.  It  is  necessary,  therefore,  to  put  point 


128  MECHANICS  [ART.  79 

No.  4  at  the  hinge.  Draw  the  lines  4-1,  1-2,  2-3,  and  3-5  parallel 
to  the  rays  ao,  bo,  co  and  do  as  in  Fig.  125.  Through  d,  on  the 
force  diagram,  draw  a  line  parallel  to  the  direction  ,of  the  right 
reaction.  Draw  the  closing  line  4-5  and  draw  a  ray  parallel  to 
it.  This  ray  intersects  the  line  from  d  at  the  point  e.  The 
length  de  gives  the  magnitude  of  the  right  reaction.  Draw  the 
closing  line  ea  of  the  force  polygon.  This  line  gives  the  direction 
and  magnitude  of  the  hinge  reaction.  The  broken  line  through 
the  hinge  may  now  be  drawn  parallel  to  ea. 

Problems 

1.  A  horizontal  beam  12  feet  long,  weighing  20  pounds,  with  its  center  of 
mass  at  the  middle,  is  hinged  at  the  right  end  and  supported  by  a  cord  at 
the  left  end.     The  cord  makes  an  angle  of  20  degrees  to  the  left  of  the 
vertical.     The  beam  carries  a  load  of  16  pounds  downward  2  feet  from  the 
left  end,  and  a  load  of  18  pounds  20  degrees  to  the  left  of  the  vertical  down- 
ward 3  feet  from  the  right  end.     Find  the  pull  on  the  cord  and  the  direction 
and  magnitude  of  the  hinge  reaction.     Check  by  moments  and  resolutions. 

2.  Find  the  reactions  at  the  supports  of  the  truss  of  Fig.  98. 

3.  A  rectangular  frame  in  a  vertical-  plane  is  10  feet  wide  horizontally 
and  4  feet  high.     It  is  hinged  at  the  upper  left  corner  and  supported  by  a 
rope  at  the  upper  right  corner.     The  rope  makes  an  angle  of  20  degrees  to 
the  left  of  the  vertical.     The  frame  weighs  160  pounds  and  its  center  of 
mass  is  at  the  middle.     A  force  of  100  pounds  at  an  angle  of  10  degrees  to 
the  left  of  the  vertical  downward  is  applied  at  the  lower  left  corner.     A 
force  of  120  pounds  at  an  angle  of  15  degrees  to  the  left  of  the  vertical  upward 
is  applied  at  the  upper  edge  at  a  distance  of  4"fe'et  from  the  upper  right 
corner.     A  force  of  150  pounds  downward  is  applied  at  the  lower  edge  at  a 
distance  of  2  feet  from  the  lower  right  corner.     Find  the  tension  in  the  rope 
and  the  direction  and  magnitude  of  the  hinge  reaction.     Check  by  moments 
and  resolutions. 


CHAPTER  VIII 
FLEXIBLE  CORDS 

80.  The  Catenary. — A  flexible  cord  or  chain  suspended  from 
two  points  takes  a  definite  form,  depending  upon  its  length  and 
the  relative  positions  of  the  points  of  support.  If  the  chain  or 
cord  is  of  uniform  weight  per  unit  of  length,  the  curve  in  a  vertical 
plane  is  a  catenary.  The  problem  of  finding  the  equation  of  the 
catenary  is  an  important  application  of  statics. 

Figure  127  shows  a  flexible  cord  suspended  from  two  points 
A  and  B.  In  this  figure  the  points  are  at  the  same  level.  This, 
however,  is  not  necessary.  The  cord  weighs  w  pounds  per  unit 


FIG.   127. 

of  length.  The  force  in  the  cord  at  every  point  is  a  tension 
parallel  to  its  length.  At  the  lowest  point  this  tension  is  hori- 
zontal. The  horizontal  component  of  the  tension  is  the  same  at 
all  points.  In  the  formulas  which  follow,  this  horizontal  compo- 
nent is  represented  by  H . 

single  letter  c. 

H 

c;  H  =  we 


Ti- 
lt is  customary  to  represent  —'by a 


w 


129 


130  MECHANICS  [ART.  81 

The  constant  c  is  the  length  of  cord  whose  weight  is  equal  to  the 
horizontal  tension  (Fig.  128). 

In  Fig.  127,  a  portion  of  cord  of  length  s  is  taken  as  the  free 
body.  The  left  end  of  the  portion  is  at  the  lowest  point  of  the 
curve,  where  the  tension  is  horizontal.  Figure  127,  II,  shows 
this  portion  of  the  cord  detached  from  the  remainder.  Three 
forces  act  on  the  portion.  The  horizontal  tension  H  is  toward 
the  left  at  the  lower  end.  The  tension  T  in  the  direction  of  the 
tangent  acts  at  the  upper  right  end.  The  weight  of  the  portion 
of  cord  is  ws  and  acts  vertically  downward.  Figure  127,  III, 
gives  the  force  triangle  for  these  forces.  In  Figure  127,  I,  ah 
element  of  length  ds  is  shown  at  the  right  end  of  the  length  s. 
The  components  of  the  element  are  dx  and  dy.  Figure  127,  IV, 
shows  the  same  element  enlarged. 

Since  the  tension  at  the  end  of  the  cord  is  along  the  tangent  at 
that  position,  T  is  parallel  to  ds  and  the  force  triangle  of  Fig. 
127,  III,  is  similar  to  Fig.  127,  IV.  These  triangles  furnish 
relations  necessary  for  finding  the  equations  of  the  catenary. 

81.  Deflections  in  Terms  of  the  Length.  —  From  the  triangles 
of  Fig.  127, 

dy        ws 
ds  =    ~T' 

This  equation  involves  three  variables,  T,  s,  and  y.     The  tension 
may  be  eliminated  by  substituting 


T  =  Vs2  +  #2,  (2) 

Equation  (1)  becomes 


ws 


ds  ==  Vw2s2  -h  w2c2      Vs2  +  c2' 
Separating  the  variables  in  Equation  (3), 

sds 

dy  =  vWT^'  (4) 

Integrating, 


y  =  V*2  +  c2  +  Ki,  (5) 


in  which  KI  is  an  arbitrary  constant  of  integration.  If  the  origin 
is  taken  at  the  lowest  point  of  the  curve,  so  that?/  =  0  when  s  =0, 
the  value  of  KI  is  found  to  be  equal  to  —  c.  It  is  customary  to 
take  the  origin  at  a  distance  c  below  the  lowest  point  of  the  curve, 
so  that  y  =  c  when  s  —  0.  When  these  values  are  substituted 


CHAP.  VIII]  FLEXIBLE  CORDS  131 

in  Equation  (5),  then  KI  is  found  to  be  equal  to  0,  and  Equation 

(5)  reduces  to 

y  =  Vs*  +  c2,  (6) 

2/2  =  s2  +  c2  Formula    VII 

A  point  on  the  curve  at  a  distance  s  from  the  lowest  point, 
measured  along  the  curve,  is  at  a  vertical  distance  y  —  c  above  the 
lowest  point  of  the  curve. 

A  right  triangle  may  be  drawn  for  which  y  is  the  hypotenuse 
-aftd  s  and  c  are  the  sides.  In  Fig.  127,  III,  the  altitude  is  ws  and 
the  base  is  H,  which  is  equal  to  we. 

T*  =  wzs*  +  w2c2  =  wz(s2  +  c2)  =  w2y2,  (7) 

T   =  wy.  (8) 

These  relations  are  shown  in  Fig.  127,  V.  From  this  figure,  it 

may  be  shown  that 

y  =  c  sec  6,  (9) 

s  =  ysmB.  (10) 

Example 

A  rope  100  feet  long  is  .stretched  between  points  at  the  same  level.  The 
rope  weighs  0.2  pound  per  foot  and  the  horizontal  component  of  the  tension 
is  50  pounds.  Find  the  sag  at  the  middle. 

Since  the  supports  are  at  the  same  level,  the  lowest  point  is  at  the  middle. 
To  find  the  elevation  of  the  ends  above  the  middle,  the  value  of  s  is  taken  as 
one-half  the  length.  \ 

s  =  50  ft., 

c  =  ~  =  250ft., 

y*    =    5Q2   +  2502, 

y  =  50\/26  =  254,95  ft. 
sag  =  y  -  c  =  254.95  -  250  =  4.95  ft. 

Problems 

1.  A  chain  weighing  240  pounds  is  120  feet  long,  and  is  stretched  between 
points  at  the  same  level.     The  horizontal  tension  is  400  pounds.     Find  the 
sag  and  the  resultant  tension  at  the  ends. 

Ans.     Sag  =  8.806  ft.,  T  =  417.6  Ib. 

2.  In  Problem  1,  find  the  sag  below  the  support  of  a  point  on  the  chain 
which  is  20  feet  from  one  support.  Ans,     4.845  ft. 

3.  A  chain  100  feet  in  length,  weighing  400  pounds,  is  stretched  between 
points  at  the  same  level.     The  resultant  tension  is  520  pounds  at  either  end. 
Find  the  horizontal  tension  and  the  sag.  Ans.     Sag  =  10  ft. 


132  MECHANICS  [ART.  82 

4.  A  rope  100  feet  in  length  is  stretched  between  points  at  the  same 
level  and  sags  10  feet  at  the  middle.  Find  the  length  of  rope  whose  weight 
is  equal  to  the  horizontal  tension,  and  find  the  slope  of  the  rope  at  the  ends. 
From  Formula  VII, 


100' 


y  +  c  =  250, 
2c  =  240,  c  =  120  ft. 

frrj 

slope  =   ™  =  0.4167. 


6.  A  flexible  rope  has  one  end  attached  to  a  fixed 
point.     The  rope  runs  over  a  smooth   pulley.     At 
F       128  *ne  uPPer  Pomt  °f  contact  wth  the  pulley,  the  rope 

is  horizontal  and  is   10  feet  below  the  fixed  point. 

The  free  end  of  the  rope  hangs  vertically  downward  a  distance  of  100  feet 
(Fig.  128).     Find  total  length.  Ans.     Total  length  =  145.8  ft. 

T        y 
6.  Show  that  jr  =  -' 

82.  Deflection  in  Terms  of  the  Span. — From  the  triangles  of 

Fig.  127, 

dy        ws        s 
dx  -:  ~H    :=  c' 

Substituting  s  =  \/y2  —  c2  to  reduce  the  variables  to  two, 

dx  c 

dy—      _  d?.  ,QX 

=^  =:    c  (6) 


Integrating  Equation  (3), 

log  (y  +  Vy2  -  c2)  =  ^  +  K2.  (4) 

C  •. 

It  has  already  been  assumed  that  y  =  c  at  the  lowest  point.  If 
the  Y  axis  is  taken  through  the  lowest  point,  then  x  =  0  when 
y  =  c.  Substituting  in  Equation  (4), 

log  c  =  K2, 
and  the  equation  becomes, 


log^ 

c  c  c 

II    -4-    -\A/2    _    /.2  » 

(6) 


CHAP.  VIII]  FLEXIBLE  CORDS  133 

2x  x 

2/2  _  C2  _  C2ec  _  2cyec  +  y2;  (8) 

X  X 

c          c  c 

y  =  |  (e    +  e     ).  Formula  VIII 

Formula  VIII  is  the  principal  equation  of  the  catenary. 
Differentiating  Formula  VIII, 


From  Equations  (1)  and  (9), 

c     *         -* 
*  =  o  (gc  ~~  e    c)  *  Formula  IX 

Zi 

The  student  who  is  familiar  with  hyperbolic  functions  will 

X  _X  X  _X 

pC         \         p         C  r*  pC     __      p          C 

recognize  -  —  --  -  as  the  hyperbolic  cosine  of  -  and  --  ~  --  as 
Z  c  Z 

the    hyperbolic    sine    of    —     Formula    VIII    may  be   written, 

y    '  or 

y  =  c  cosh  -;  Formula  IX  may  be  written  s  =  c  sinh  -.     Where 
c  c 

tables  of  these  functions  are  available,  the  labor  of  solving  some 
of  the  following  problems  is  materially  reduced. 

Example 

A  rope  weighing  0.4  pounds  per  foot  is  stretched  between  points  100  feet 
apart  at  the  same  level.  The  horizontal  tension  is  120  pounds.  Find  the 
sag  and  the  length  of  the  rope. 

This  is  a  case  where  x  and  c  are  given.  For  the  entire  rope,  x  =  50  and 
c  =  300. 

y  =  I50(e*  +  e'1); 
loglo/=  0,4342945  =0()723824; 

D 

e*=  1.1813. 
logloe"¥  =  1.9276176; 

e~*  =  0.8465 

y  =  150  X  2.0278  =  304.17ft. 
y  -  c  =  4.17ft. 

If  a  table  of  hyperbolic  functions  were  available,  the  solution  would  be 
y  =  150  cosh  0.16667. 


134  MECHANICS  [ART.  82 

To  find  the  length  of  half  the  rope,  Formula  VII  might  now  be  used. 
Since  the  values  of  e*  and  e~*  have  already  been  found,  Formula  IX  is  more 
convenient. 

s  =  150(1.1813  -  0.8465)  =  50.02  ft. 


Problems 

1.  A  cable  weighing  1  pound  per  foot  is  stretched  between  points  400 
feet  apart  at  the  same  level.     The  horizontal  tension  is  800  pounds.     Find 
the  sag  at  the  middle,  the  length  of  the  cable,  and  the  direction  and  magni- 
tude of  the  resultant  tension  at  each  end. 

2.  A  chain  100  feet  in  length  is  stretched  between  points  at  the  same 
level  and  sags  5  feet.     Find  the  horizontal  distance  between  the  points. 

Solve  first  for  c  and  y,  as  in  Problem  4  of  the  preceding  article,     c  = 
247.5  ft.,  y  =  252.5  ft. 


_  c  I  *      -^\ 

S~2\ec~e    c)> 


y  -f  s  =  c 
which  is  the  same  as  equation  (7). 


g  _  302.5          302.5  _  x 
ec  ~  247^  '     ge  247.5  ~  c' 

=  0.08715,  loge  =  °-200669- 


2x  =  0.200669  X  2  X  247.5  =  99.33  ft. 

The  second  form  of  Equation  (5)  gives  the  same  method. 

3.  A  steel  tape  100  feet  long  is  stretched  between  points  at  the  same  level 
and  sags  2  feet.     What  is  the  horizontal  distance  between  the  points?     If 
the  tape  weighs  1  pound,  what  is  the  horizontal  tension? 

Ans.     2x  =  99.89  ft.;  H  =  6.24  Ib. 

4.  Solve  Problem  3  if  the  sag  is  only  1  foot. 

6.  A  steel  tape  100  feet  long,  weighing  1  pound,  is  stretched  between 
points  at  the  same  level.  The  horizontal  tension  is  12  pounds.  Find  the 
distance  between  the  points. 

In  this  problem,  s  is  given  and  x  is  to  be  found.  Formula  VII  might 
be  used  to  find  y,  and  this  value  of  y  might  be  substituted  in  Equation  (6). 
Another  method  is  by  Formula  IX. 

X  _X 

1  =  12(e<  -  e    c). 

X 

Multiplying  by  ec 

x  2x 

e°  =  12ec   -  12, 
2x         x 
I2e  c   -  ec  -  12  =  0. 


CHAP.  VIII]  FLEXIBLE  CORDS  135 

Solving  this  quadratic, 

X 

e~c  =  1.042534, 
from  which 

2x  =  99.96  ft. 

6.  A  chain  120  feet  long,  weighing  100  pounds,  is  stretched  between 
points  at  the  same  level.  The  resultant  tension  at  either  end  is  130  pounds. 
Find  the  horizontal  distance  between  the  points  and  the  sag  at  the  middle. 

83.  Solution  by  Infinite  Series.  —  In  the  problems  of  the  pre- 
ceding articles,  the  values  of  c  were  easily  calculated  from  the 
tension  and  weight  per  foot  or  from  the  length  and  the  sag. 
Problems  for  which  c  can  not  be  so  calculated  are  more  difficult 
to  solve. 

Example 

A  rope  is  stretched  between  points  at  the  same  level  100  feet  apart  and 
sags  4  feet.  Find  the  tension  in  terms  of  the  weight  of  1  foot  of  rope. 

50  _50 

°  +e 

50  _50 

c    +  e     °]    ~c,  (2) 

50  _50 

8  =  c  (e  °   +  e     c    -  2\.  (3) 

Equation  (3)  may  be  solved  by  the  method  of  trial  and  error.  The  labor 
is  much  reduced  if  a  suitable  table  of  hyperbolic  cosines  is  available.  An 
approximate  value  of  c  may  easily  be  obtained.  Where  the  sag  is  small,  s 
is  very  nearly  equal  to  x.  Taking  s  =  50,  and  y  —  c  =  4  the  value  of  c  is 
found  to  be  310.5  ft.  Substituting  in  Equation  (3), 

8  =  310.5(1.17472  +  0.85126  -  2)  =  8.067. 

A  closer  approximation  may  be  obtained  by  assuming  that  s2  =  2500  +  16. 
This  assumption  gives  c  =  312.5  ft.  When  this  value  of  c  is  substituted  in 
Equation  (3),  the  result  is 

8  =  8.015. 
Exterpolating,  c  =  313.1  ft. 

To  get  an  algebraic  expression  for  s  in  terms  of  x  and  c,  expand 


into  a  series  by  means  of  Maclaurin's  Theorem.     This  theorem  is 

/(*)  =  /(O)  +  f(0)x  +  /"(0)|  +  /'"(0)|  +  , 
in  which  /  (x)  is  the  quantity  required,  /(O)  of  is  the  value  f(x) 


136  MECHANICS  [ART.  83 

when  x  becomes  0,  /'(0)is  the  value  of  the  first  derivative  of 
f(x)  when  x  becomes  0,  etc. 

Expand  ex  —  e~x  by  Maclaurin's  Theorem. 

f(x)  =e*-e~*  ,/(0)  =0; 

/'(*)  =e*+e-*,  f(0)  =  2; 

/"(*)  =  e*-e-*,/"(0)  =0; 

/'"(*)  =  e*+e-*,f'"(Q)  =2. 


^r+0+-M-  +  etc. 
3  51 


-  e-x 
--  =  smh 


Equation  (4)  is  the  same  as  the  series  for  sin  x,  except  that 
all  the  terms  are  positive.  In  the  series  for  sin  x,  the  terms  are 
alternately  positive  and  negative. 

or 

Substituting  -  for  x  and  multiplying  by  c, 
c 

X  _X 

ec  _  e     c 


In  a  similar  manner, 


X  _X 

"c  -L  e~~c  I  X2  X*  \ 

—^~-  =c(l+  272+4574  +  etc.  j;  (7) 


When  x  is  small  compared  with  c,  as  is  the  condition  in  cables 
with  small  sag,  these  series  converge  rapidly  and  two  terms  are 
generally  sufficient  to  obtain  a  fairly  accurate  result. 


Problems 

1.  A  rope  weighing  0.2  pounds  per  foot  is  stretched  between  points  at 
the  same  level  which  are  160  feet  apart.  The  horizontal  tension  is  80 
pounds.  Find  the  sag  at  the  middle  by  one  term  of  Equation  (8).  Solve 
also  by  means  of  two  terms  of  Equation  (8).  Compare  the  results  with 
those  obtained  by  Formula  VIII.  Ans.  8  ft.;  8.026  ft.;  8.026  ft. 


CHAP.  VIII] 


FLEXIBLE  CORDS 


137 


2.  Find  the  length  of  the  rope  of  Problem  1  by  the  series  and  by  Formula 
IX.  Ans.     161.07ft.;  161.11  ft. 

3.  A  chain  102  feet  long  is  stretched  between  points  100  feet  apart  at 
the  same  level.     Find  the  horizontal  tension  in  terms  of  the  weight  of  unit 
length  of  chain. 

In  this  problem  s  and  x  are  given  and  c  is  unknown.  A  problem  of  this 
kind  can  not  be  solved  directly  by  the  exponential  equations  of  the  catenary. 
If  two  terms  of  the  series  of  Equation  (6)  are  used, 

i.<B-i+££ 

from  which  c  =  144.3  ft.     If  three  terms  of  the  series  are  used, 


from  which  c  =  144.7  ft. 

4.  Solve  the  example  at  the  beginning  of  this  article  by  means  of  one 
term  of  Equation  (9). 

5.  A  chain  weighing  1  pound  per  foot  is  stretched  between  points  120 
feet  apart  at  the  same  level.     Find  the  sag  at  the  middle  when  the  horizontal 
tension  is  180  pounds.     Solve  by  one  term  of  the  series  and  by  Formula 
VIII. 

6.  Solve  Problem  5  if  the  horizontal  tension  is  60  pounds. 

84.  Cable  Uniformly  Loaded  per  Unit  of  Horizontal  Distance. 
A  suspension  bridge   consists  of  trusses  supported  by  cables. 


\wx 


III 


H 
FIG.  129. 

The  weight  of  each  truss  and  the  part  of  the  floor  which  it  carries 
is  much  greater  than  that  of  the  supporting  cable.  The  load 
on  the  cable  is  nearly  constant  for  the  unit  of  horizontal  distance. 
If  w'  is  the  weight  per  unit  length  of  the  combined  truss  and 
cable,  the  total  weight  in  length  x  is  w'x.  Figure  129,  II,  shows  a 
part  of  such  a  cable.  The  distance  x  is  measured  from  the  lowest 


138  MECHANICS  [ART.  84 

point.  The  forces  acting  on  the  portion  of  cable  are  the  hori- 
zontal tension  H,  the  tension  T  in  the  direction  of  the  tangent  at 
the  other  end  of  the  portion,  and  the  vertical  load  w'x.  Figure 
129,  III,  is  the  force  triangle. 

Tfln  fi  _dy  _w'x  _x 

T™e-dx~^r  -?' 

.,  H 

lf  w'  =  c 

,         xdx  /n. 

dy  =  -j-  ;  (2) 

T*2 

...          »-!?+*•      ..     . 

If  the  origin  of  coordinates  is  taken  at  the  lowest  point,  y  =  0 
when  x  =  0,  and  K  =  0. 


This  is  the  equation  of  a  parabola  with  the  axis  vertical. 
To  find  the  length  of  this  parabola, 


Expanding  Equation  (4)  by  the  binomial  theorem, 

/-•  X*  X*  X*  5#8     \  J  /r\ 

ds  ==  I1  +2^  -  M*  +  167°  ~  IW)  ** 
Integrating  Equation  (5) 

xs          x5  x7  5x9 

f     /2~/4  +         /6  "1 


A  comparison  of  Equation  (6)  for  the  parabola  with  Equa- 
tion (6)  for  the  catenary  shows  that  the  first  two  terms  of  the 
series  are  identical  if  c  =  c1  '.  Comparison  of  Equation  (3)  for 
the  parabola  with  Equation  (9)  for  the  catenary  shows  that  the 
parabola  is  equivalent  to  the  first  term  of  the  series  for  the 
catenary.  There  are  few  problems  to  which  the  parabola  formula 
strictly  applies.  Engineers  use  it  frequently  as  an  approxima- 
tion for  the  catenary.  When  so  used,  it  is  better  to  consider  it 
as  the  first  term  of  the  series  of  Equation  (9)  of  Art.  83,  rather 
than  to  assume  that  the  load  is  uniformly  distributed  per  unit  of 
horizontal  distance. 


CHAP.  VIII]  FLEXIBLE  CORDS  139 

Do  not  make  the  common  mistake  of  using  an  approximate 
formula  for  a  problem,  when  it  is  easy  to  solve  by  the  correct 
catenary  equations. 

Problems 

1.  A  steel  cable  weighing  1  pound  per  foot  carries  a  lead-sheathed  tele- 
phone cable  weighing  3  pounds  per  foot.     The  lead  cable  is  suspended  from 
the  steel  cable  in  such  a  way  that  it  is  practically  horizontal.     (This  is  not 
generally  done  with  telephone  cables.)     Assuming  that  the  combined  weight 
per  horizontal  foot  is  constant,  find  the  tension  in  the  steel  cable  if  the  span 
is  400  feet  and  the  sag  is  20  feet. 

Ans.     cf  =  1000  ft.;  T  =  4080  Ib. 

2.  What  must  be  the  height  of  the  towers  to  carry  cables  such  as  those  of 
Problem  1  across  a  river  800  feet  wide  if  the  maximum  tension  is  5000 
pounds  and  if  the  cable  must  ber  at  least  40  feet  above  the  water? 

85.  Summary. — In  the  case  of  a  flexible  cord  or  chain  with 
uniform  load  per  unit  of  length,  the  following  cases  are  important : 

(1)  Span     and     Horizontal     Tension    Given. — Substitute    in 
Formula  VIII  for  the  sag  and  in  Formula  IX  for  the  length. 
Check  by  Formula  VII. 

For  an  approximate  value  of  the  sag  use  Equation  (9)  of  Art. 
83.  The  first  term  of  the  series  is  equivalent  to  the  equation  for 
a  cord  with  uniform  load  per  unit  of  horizontal  distance,  and 
gives  a  fair  approximation  when  the  sag  is  small  compared  with 
the  length.  For  more  accurate  results  use  two  or  three  terms. 
For  an  approximate  value  of  the  length,  use  Equation  (6)  of 
Art.  83. 

(2)  Length  and  Sag  Given. — Solve  for  c  by  Formula  VII,  and 
then  for  x  by  Equation  (5)  of  Art.  82,  or  by  Formula  VIII. 

(3)  Length    and    Horizontal    Tension   Given. — Substitute    the 

X 

values  of  s  and  c  in  Formula  IX.     Multiply  by  ec  to  form  a 

X 

quadratic  in  which  the  unknown  is  ec.     Solve  for  y  by  Formula 
VII. 

X 

The  term  e~  c  may  be  eliminated  by  adding  Formulas  VIII  and 

X 

IX  instead  of  by  multiplying  by  ec. 

(4)  Length  and  Total  Tension  Given. — Solve  for  the  horizontal 
tension  by  means  of  the  force  triangle;  then  use  the  methods  of 
Case  (3)  above. 

(5)  Length  and  Span  Given. — In  this  case  c  is  unknown  and 
Formula  VII  does  not  apply.     Use  the  series  of  Art.  83  as  in 
Problem  3  of  that  article. 


140  MECHANICS  [ART.     86 

(6)  Sag  and  Span  Given. — Use  Equation  (9)  of  Art.  83  to  get 
the  value  of  c.  The  first  term  of  the  series  is  the  same  as  the 
equation  of  a  cord  with  uniform  load  per  unit  horizontal  length. 
The  use  of  the  next  term  involves  the  solution  of  a  cubic. 

After  c  has  been  found,  the  length  may  be  computed  by  For- 
mula VII  or  by  the  series  for  s. 

86.  Miscellaneous  Problems 

1.  A  flexible  cable,  weighing  1  pound  per  foot,  is  stretched  across  a  river 
1200  feet  wide.     The  allowable  tension  is  6000  pounds.     What  must  be 
the  height  of  the  supports  if  the  cable  is  not  permitted  to  come  within  60 
feet  of  the  water?     Solve  as  a  catenary. 

2.  A  flexible  cord  200  feet  long,  is  stretched  between  two  points  of  support. 
The  lowest  point  of  the  cord  is  8  feet  below  one  of  the  points  and  12  feet 
below  the  other.     Find  the  horizontal  distance  between  the  supports. 

Ans.  Lowest  point  is  89.8  ft.  from  the  lower  support,  measured  along 
the  cord.  The  total  horizontal  distance  between  supports  =  198.66  ft. 

3.  A  cable  across  a  river  800  feet  in  width  is  used  to  supply  concrete  for 
a  dam.     The  cable  weighs  2  pounds  per  foot.     The  allowable  tension  is 
10,000  pounds.     The  maximum  load  is  1200  pounds.     Find  the  sag  at  the 
middle  when  this  load  is  at  the  middle.     Solve  by  approximate  formula. 
Consider  the  additional  load  as  replaced  by  additional  cable  in  the  middle. 

Ans.     Sag  at  middle  =  40.4  ft.  with  no  allowance  for  stretch  of  the  cable. 


CHAPTER  IX 
CONCURRENT  NON-COPLANAR  FORCES 

87.  Resultant  and  Components. — It  is  an  experimental  fact 
that  the  resultant  of  two  concurrent,  coplanar  forces  is  repre- 
sented by  their  vector  sum.     Starting  from  this  fact,  it  has  been 
shown  that  the  resultant  of  any  number  of  concurrent,  coplanar 
forces  may  be  represented  by  their  vector  sum.     Finally,  it  has 
been  proved  that  the  direction  and  magnitude  of  the  resultant 
of  any  number  of  non-concurrent,  coplanar  forces  may  be  deter- 
mined from  the  vector  sum  of  these  forces. 

The  next  problem  for  consideration  is  that  of  concurrent,  non- 
coplanar  forces.  Figure  130  shows  two  forces,  P  and  Q  in  a  verti- 
cal plane,  and  a  single  force  H  in  the  horizontal  plane.  The 
resultant  of  P  and  Q,  or  of  any  number  of  forces  in  the  same 
plane,  is  the  single  force  R  in  that  plane.  This  resultant  R  and 
the  force  H  intersect  at  the  point  of  application.  They  lie  in 
the  inclined  plane  OAB  of 
Fig.  130,  II.  Since  R  and  H 
are  in  a  plane,  their  resultant 
is  their  vector  sum  in  that 
plane.  The  force  H  as  well 
as  the  force  R  may  be  the 
resultant  of  several  forces  in 
its  plane.  In  Fig.  130  the 
original  planes  are  normal  to 
each  other.  There  is  nothing 
in  the  proof  which  is  limited  to  this  position  of  the  planes.  They 
may  have  any  angle  with  each  other. 

The  resultant  of  concurrent  forces  in  space  may  be  represented  by 
their  vector  sum.  The  forces  which  make  up  the  resultant  are  its 
components. 

88.  Resolution  of  Non-coplanar  Forces. — The  most  important 
components  are  those  obtained  by  orthographic  projection  upon  a 
line  or  upon  a  plane.     If  the  force  is  projected  upon  three  lines 
at  right  angles  to  each  other,  the  three  components  form  the 
edges  of  a  rectangular  parallelepiped.     The  resultant  of  these 
three  components  is  the  diagonal  of  the  parallelepiped. 

141 


142 


MECHANICS 


[ART.   89 


Example 

A  cord  is  attached  to  a  fixed  point  A,  passes  over  a  pulley  B,  and  carries 
a  load  of  24  pounds  on  the  free  end.  The  pulley  is  4  feet  south  and  3  feet 
east  of  a  point  C.  The  point  C  is  6  feet  above  the  point  A.  Find  the 
components  of  the  pull  in  the  cord  vertically  upward,  horizontally  toward 
the  south,  and  horizontally  toward  the  east, 

The  length  of  the  diagonal  from  A  to  B  is  A/61  ft.     East  component  = 

3  4 

X  24  =  9.22  Ib.     South  component  =  —7=  X  24  =  12.29  Ib.     Ver- 


V61 

tical  component  =  18.44  Ib. 


South  component  =  —/=  X  24  =  12.29   Ib. 


Horizontal  Com, 


I 

-I—  I  ---- 


It  will  be  observed  that  the  components 
are  proportional  to  the  edges  of  a  rec- 
tangular parallelepiped  of  which  the 
dimensions  are  3  ft.,  4  ft.,  and  6  ft. 

Figure  131  gives  the  graphical  solution 
for  these  components. 

The  student  should  make  a  sketch  of 
the  cord  and  pulley. 

Problems 


Force 


1.  A  horizontal  trap-door,  6  feet  by  10 
feet  is  lifted  by  a  rope  attached  to  one 
FIG.  131.  corner.     The  rope  passes  over  a  pulley, 

which  is  7  feet  above  the  middle  of  the 

door,  and  carries  a  load  of  40  pounds  on  the  free  end.  Find  the  vertical 
component  of  the  pull  on  the  door  and  find  the  horizontal  components 
parallel  to  the  edges.  Ans.  Vertical  component  =  30.73  Ib. 

2.  Solve  Problem  1  graphically,  using  the  scale  of  1  inch  =  2  feet  in  the 
space  diagram,  and  1  inch  =  10  pounds  in  the  force  diagram. 

3.  A  pull  of  50  pounds  is  applied  by  a  rope  which  makes  an  angle  of  35 
degrees  with  the  vertical  in  a  vertical  plane  wh'ch  is  north  20  degrees  east. 
Find  the  vertical  component  of  the  force  and  the  horizontal  components 
east  and  north.  Ans.     Horizontal  component  north  =  26.95  Ib. 

4.  Solve  Problem  3  graphically  to  the  scale  of  1  inch  =  10  pounds. 

6.  A  force  of  20  pounds  makes  an  angle  of  42  degrees  with  the  vertical, 
and  an  angle  of  55  degrees  with  the  horizontal  line  toward  the  north.  Find 
its  vertical  component  and  its  horizontal  components  east  and  north. 

Ans.     Horizontal  component  east  =  6.89  Ib. 

6.  Solve  Problem  5  graphically  to  the  scale  of  1  inch  =  5  pounds. 

89.  Calculation  of  Resultant. — To  find  the  resultant  of  a 
number  of  non-coplanar  forces,  it  is  best  to  resolve  each  force 
along  each  of  three  axes  at  right  angles  to  each  other,  to  add  the 
components  along  each  axis  to  get  the  component  of  the  resultant 
along  that  axis,  and  finally  to  combine  these  components  to  form 
a  single  force.  The  components  of  the  resultant  are  the  edges 
of  a  rectangular  parallelepiped  of  which  the  resultant  is  the 


CHAP.  IX]    CONCURRENT  NON-COPLANAR  FORCES    143 

diagonal.  The  resolutions  and  compositions  may  be  made 
algebraically  or  graphically.  In  the  graphical  method  it  is  best 
to  resolve  each  force  on  a  separate  diagram. 

Example 

Find  the  resultant  of  the  following  three  forces:  10  pounds  at  38  degrees 
with  the  vertical  in  a  vertical  plane  which  is  north  25  degrees  east;  12  pounds 
east  of  north  at  an  angle  of  45  degrees  with  the  vertical  and  an  angle  of  60 
degrees  with  the  north  horizontal;  and  13  pounds  applied  by  means  of  a 
rope  which  passes  over  a  pulley  4  feet  east,  3  feet  south,  and  12  feet  above 
the  point  of  application  of  the  forces. 


Force 

Components 

Up 

North 

East 

10 

7.880 

5.580 

2.602 

12 

8.484 

6.000 

6.000 

13 

12.000 

-3.000 

4.000 

28.364 

8.580 

12.602 

If  0  is  the  angle  which  the  vertical  plane  of  the  resultant  makes  with  the 
north  vertical  plane, 

=  12.602 

0  =  55°  45'. 

If  H  is  the  resultant  of  the  two  horizontal  components, 

12.602 


sin  0 


=  15.25  Ib. 


If  6  is  the  angle  which  the  resultant  makes  with  the  vertical, 
tan0  = 


R 


28.36 
28.36 
cos  0 


,  0  =  28°  16'. 
32.20  Ib. 


The  resultant  is  a  force  of  32.20  pounds  at  an  angle  of  28°  16'  with  the 
vertical  in  a  vertical  plane  which  is  north  55°  45'  east. 

Problems 

1.  Find  the  resultant  of  25  pounds  at  an  angle  of  42  degrees  with  the 
vertical  in  a  vertical  plane  which  is  north  34  degrees  east,  and  20  pounds 
at  an  angle  of  56  degrees  with  the  vertical  in  a  vertical  plane  which  is  north 
20  degrees  west.  Solve  by  resolutions  along  east  and  north  horizontal  axes, 


144 


MECHANICS 


[ART.   90 


and  also  solve  with  one  horizontal  axis  north  70  degrees  east  and  the  other 
north  20  degrees  west. 

2.  Solve  Problem   1   graphically  to  the  scale  of   1  inch  =  10  pounds. 
Make  a  separate  diagram  for  each  resolution,  and  a  third  diagram  for  the 
composition. 

3.  Find  the  resultant  of  the  following  forces:  40  pounds  east  of  north 
at  an  angle  of  50  degrees  with  the  vertical  and  60  degrees  with  the  hori- 
zontal north;  50  pounds  east  of  south,  at  an  angle  of  70  degrees  with  the 
vertical  and  65  degrees  with  the  horizontal  south;  40  pounds  above  the 
horizontal  at  an  angle  of  62  degrees  with  the  horizontal  north  and  40  degrees 
with  the  horizontal  east. 

Ans.  115.4  Ib.  at  an  angle  of  57°  37'  with  the  vertical,  in  a  vertical  plane 
north  79°  34'  east. 

90.  Equilibrium  of  Concurrent,  Non-coplanar  Forces. — The 

condition  for  equilibrium  of  concurrent,  non-coplanar  forces  is 
the  same  as  for  coplanar  forces.  The  force  polygon  must  close, 
or  the  resultant  of  all  the  forces  must  equal  zero.  Since  the 
force  polygon  is  not  all  in  one  plane,  three  components  are 
required  to  specify  fully  some  of  the  forces.  There  are  three 
unknowns  and  three  independent  equations.  The  sum  of  the 
components  along  any  axis  is  zero.  In  order  that  the  equations 
may  be  independent,  one  of  these  axes  must  have  a  component 
perpendicular  to  the  plane  of  the  other  two. 

The  principal  difficulty  in  the  solution  of  a  problem  of  con- 
current, non-coplanar  forces  is  that  of  finding  the  required  angles 
from  the  conditions  of  the  problem. 

Example 

A  mass  of  40  pounds  is  supported  by  three 
ropes.  The  ropes  are  fastened  to  a  horizontal 
ceiling  8  feet  above  the  body  at  points  A,  B, 
and  C,  respectively.  If  E  is  a  point  in  the 
ceiling  directly  over  the  body,  the  point  A  is  3 
feet  north  and  4  feet  east  of  E;  the  point  B  is  3 
feet  north  and  4  feet  west  of  E',  the  point  Cis 
6  feet  directly  south  of  E.  Find  the  tension  in 
each  rope. 

A  horizontal  resolution  perpendicular  to  the 
vertical  plane  ECO  of  Fig.  132  eliminates  the 
force  in  OC.  The  ropes  OA  and  OB  make  equal 
angles  with  the  axis  of  resolution.  If  the  forces  in  the  ropes  are  represented 
by  A,  B,  and  C,  respectively, 

A  =  B. 
By  a  vertical  resolution, 

2A  X  — 7^  +  0.8C  =  40. 


40 


Ib. 


FIG.  132. 


' 


CHAP.  IX]    CONCURRENT  NON-COPLANAR  FORCES    145 


By  a  horizontal  resolution  parallel  to  EC 

3 


2A  X 


A/89 

5A/89 


-  0.6C  =  0; 
=  15.72  Ib. 


C   =  16.67  Ib. 


Problems 

1.  In  the  example  above,  let  B  be  3  feet  west  of  the  plane  ECO',  let  the 
other  data  remain  unchanged.     Find  the  tension  in  each  rope. 

Ans.     C  =  16.67  Ib.;  A  =  13.48  Ib.;  B  =  17.34  Ib. 

2.  A  40-pound  mass  is  supported  by  three  ropes      Each  rope  is  10  feet 
long.     The  ends  of  the  ropes  are  attached  to  a  horizontal  ceiling  at  the 
vertices  of  an  equilateral  triangle  each  side  of  which  is  8  feet.     Find  the 
tension  in  each  rope. 

91.  Moment  about  an  Axis. — The  moment  of  a  force  about  an 
axis  is  the  product  of  the  component  of  the  force  in  a  plane  per- 
pendicular to  the  axis  multiplied  by  the  shortest  distance  from 
the  line  of  the  force  to  the  axis.  In  Fig.  133,  the  force  is  P 
and  the  axis  is  OZ.  The  XY  plane  is  perpendicular  to  the  axis. 
The  force  P  may  be  resolved  into  two  components  at  right 
angles  to  each  other.  One 
component,  which  is  not  shown 
in  Fig.  133,  is  parallel  to  the 
axis  OZ.  The  other  compo- 
nent is  P'  in  the  plane  perpen- 
dicular to  the  axis.  The, 
distance  d  from  the  axis  to  the 
line  of  the  component  P'  is  the 
effective  moment  arm. 


M  =  P'd. 


FIG.   133. 


In  Art.  47,  the  moment  of  a  force  was  said  to  be  about  a 
point  in  its  plane.  It  is  better  to  consider  the  moment  to  be 
about  an  axis  which  passes  through  the  point  and  which  is 
perpendicular  to  the  plane. 

In  the  case  of  the  moment  of  a  force  about  a  point  in  its  plane, 
the  apparent  arm  was  drawn  from  the  origin  of  moments  to 
the  line  of  the  force;  then  this  arm  was  multiplied  by  the  compo- 
nent of  the  force  perpendicular  to  it.  Likewise,  in  the  case  of  the 

moment  of  a  force  with  respect  to  any  axis,  a  plane  may  be  passed 
10 


146  MECHANICS  [ART.  92 

through  the  axis  of  moments  intersecting  the  line  of  force  at  any 
convenient  point.  The  component  of  the  force  normal  to  this 
plane  multiplied  by  the  perpendicular  distance  from  the  point  of 
intersection  to  the  axis  gives  the  required  moment. 

In  Fig.  133,  the  plane  OXZ  passes  through  the  axis  OZ.  The 
line  of  force  may  be  extended  until  it  intersects  this  plane  at 
C.  The  moment  of  the  force  about  OZ  is  the  product  of  the 
vertical  component  Pv  multiplied  by  the  distance  from  C  to  the 
axis.  The  other  components  of  the  force  P  at  this  point  lie  in 
the  plane  OXZ  and  either  intersect  the  axis  or  are  parallel  to  it. 
Consequently  only  the  vertical  component  Pv  has  moment  about 
the  axis. 

The  line  of  the  force  P  might  be  extended  in  the  opposite 
direction  till  it  intersects  the  vertical  plane  OYZ.  The  horizon- 
tal component  of  the  force  at  this  point  of  intersection  multiplied 
by  the  vertical  distance  from  the  point  to  the  axis  gives  the 
moment.  Any  plane  through  the  axis  may  be  used  in  this  way. 
The  choice  of  a  plane  depends  upon  the  difficulty  of  finding  the 
component  of  the  force  perpendicular  to  it,  and  of  finding  the 
distance  from  the  point  of  intersection  to  the  axis. 

Example 

In  Fig.  132,  find  the  moment  of  the  force  in  OC  about  the  axis  AB. 
Use  first  the  horizontal  plane  ABC.     The  moment  arm  is  9  feet  in  length. 
If  C  is  the  magnitude  of  the  force  in  OC,  its  vertical  component  is  0.8(7. 

M  =  0.8C  X  9  =  7.2C. 

Use  next  the  vertical  plane  through  the  axis.  Extend  the  line  of  CO 
beyond  0  till  it  intersects  this  plane.  The  point  of  intersection  is  12  feet 
from  B A  and  the  horizontal  component  of  the  force  is  0.6C. 

M  =  0.6C  X  12  =  7.2C. 

92.  Equilibrium  by  Moments. — If  a  body  is  in  equilibrium 
under  the  action  of  a  number  of  concurrent,  non-coplanar  forces, 
the  sum  of  the  moments  of  these  forces  about  any  axis  is  zero. 
If  a  plane  is  passed  through  the  axis  of  moments  and  the  point  of 
application  of  the  forces,  the  moment  of  each  force  will  be  the 
product  of  its  component  perpendicular  to  the  plane  multiplied 
by  the  distance  of  the  point  of  application  from  the  axis.  Since 
the  moment  arm  is  the  same  for  all  the  concurrent  forces,  the 
sum  of  the  moments  is  equivalent  to  a  resolution  perpendicular 
to  the  plane.  It  is  not  necessary  to  use  this  plane  in  calculating 


CHAP.  IX]    CONCURRENT  NON-COPLANAR  FORCES    147 


the  moments.  Any  convenient  plane  may  be  used,  and  different 
planes  may  be  employed  for  the  different  concurrent  forces. 
Instead  of  three  resolution  equations  of  equilibrium,  there  may 
be  one  moment  and  two  resolutions,  two  moments  and  one  resolu- 
tion, or  three  moments. 

Example 

Find  the  force  C  of  Fig.  132  by  moments  about  AB.  The  line  AB  is 
taken  as  the  axis  because  it  eliminates  two  unknowns.  The  moment  of  C 
about  AB  has  already  been  found  to  be  7.2  C.  The  moment  of  the  weight 
of  40  pounds  is  40  X  3  =  120  ft.-lb. 

7.2C  =  120, 
C  =  16.67  Ib. 


With  C  known,  A  and  B  may  best  be  deter- 
mined by  two  resolution  equations.  The 
method  of  moments,  however,  may  be  em- 
ployed. To  find  the  force  B  take  moments 
about  the  horizontal  line  AC.  Figure  134 
shows  the  top  view,  or  plan  of  Fig.  132.  The 
moment  arm  of  the  40-pound  mass  is  the  line 
EF  whose  length  is  6  sin  6.  The  moment 
arm  of  the  vertical  component  of  the  force 
B  is  8  cos  6.  From  the  figure,  tan  B  =  f . 


FIG.  134. 


8B 

V89 


8  cos  6  =  40  X  6  sin  0; 


B  =  15V/89  tan  6  = 


=  15.72  Ib. 


Problems 

1.  Solve  Problem  1  of  Art.  90  by  moments. 

2.  A  mass  of  60  pounds  is  suspended  by  two  ropes,  each  13  feet  in  length, 
attached  to  a  horizontal  ceiling  at  points  10  feet  apart,  and  by  a  third  rope, 
15  feet  in  length,  attached  to  the  same  ceiling  at  a  point  12  feet  from  the 
point  of  attachment  of  each  of  the  others.     Find  the  tension  in  each  rope 
by  moments  and  check  by  a  vertical  resolution. 

93.  Summary.  —  The  resultant  of  concurrent,  non-coplanar 
forces  is  represented  by  their  vector  sum. 

The  moment  of  a  force  about  an  axis  is  the  moment  of  its 

.component  in  a  plane  perpendicular  to  the  axis  about  the  point 

of  intersection  of  the  axis  with  the  plane.     To  calculate  the 

moment,  pass  a  plane  through  it  and  some  point  on  the  line 

of  the  force.     The  component  of  the  force  normal  to  the  plane 


148  MECHANICS  [ART.  93 

multiplied  by  the  distance  of  the  point  from  the  axis  gives  the 
required  moment. 

For  equilibrium  the  force  polygon  closes.  There  are  three 
unknowns  and  three  independent  equations  may  be  written. 
These  equations  may  be: 

(1)  Three  resolutions. 

(2)  One  moment  and  two  resolutions. 

(3)  Two  moments  and  one  resolution. 

(4)  Three  moments. 

A  moment  equation  for  non-coplanar  concurrent  forces  is 
equivalent  to  a  resolution  perpendicular  to  the  plane  which  passes 
through  the  axis  of  moments  and  the  point  of  application  of  the 
forces. 

In  writing  resolution  equations,  each  direction  of  resolution 
must  have  a  component  perpendicular  to  the  plane  of  the  others. 
It  is  not  necessary  that  these  directions  be  at  right  angles  to  each 
other.  A  resolution  perpendicular  to  the  direction  of  a  force 
eliminates  that  force. 

When  moment  and  resolution  equations  are  written  together, 
a  resolution  must  not  be  taken  perpendicular  to  the  plane  through 
the  axis  of  moments  and  the  point  of  application  of  the  forces. 
When  two  resolutions  and  one  moment  equation  are  written,  the 
plane  of  the  directions  of  the  two  resolutions  must  not  be  perpen- 
dicular to  the  moment  plane. 

Two  axes  of  moments  and  the  point  of  application  of  the  forces 
must  not  lie  in  the  same  plane.  Three  axes  of  moment  must  not 
be  parallel. 


CHAPTER  X 


PARALLEL  FORCES  AND  CENTER  OF  GRAVITY 

94.  Resultant  of  Parallel  Forces.  —  Articles  55  and  56  show  that 
the  resultant  of  two  or  more  parallel  forces  in  the  same  plane 
is  equal  to  their  algebraic  sum,  and  that  the  sum  of  the  moments 
of  all  the  forces  about  any  point  in  their  plane,  is  equal  to  the 
moment  of  their  resultant  about  that  point.  It  will  now  be 
shown  that  the  resultant  of  parallel  forces  which  are  not  all  in  the 
same  plane  is  the  algebraic  sum  of  the  forces,  and  that  the  sum  of  the 
moments  of  all  the  forces  about  any  axis  is  equal  to  the  moment  of 
their  resultant  about  that  axis. 

Figure  135  shows  two  forces,  P  and  Q  together  with  their 
resultant  R.  The  moment  of  R  about  any  point  in  their  plane 


II 


FIG.  135. 

is  equal  to  the  sum  of  the  moments  of  P  and  Q  about  that  point. 
The  origin  of  moments  at  the  point  0  may  be  regarded  as  the 
point  at  which  an  axis  perpendicular  to  the  plane  of  the  forces 
intersects  that  plane.  In  Fig.  135,  II,  the  axis  is  OZ.  If  the 
moment  of  the  resultant  is  equal  to  the  sum  of  the  moments  of 
the  two  forces  about  an  axis  perpendicular  to  their  plane,  the 
moments  of  these  forces  about  any  other  axis  through  0  are 
likewise  equal.  To  change  from  the  moment  about  the  perpen- 
dicular axis  to  the  moment  about  an  inclined  axis,  it  is  necessary 
to  multiply  each  term  by  the  cosine  of  the  same  angle;  hence,  if 
the  moments  are  equal  in  the  one  case,  they  are  equal  in  the  other. 
Let  S  be  a  third  force  which  is  parallel  to  P  and  Q  but  is  not 
in  the  same  plane.  A  plane  may  be  passed  through  R  and  S.  To 

149 


150  MECHANICS  [ART.  95 

find  their  resultant  the  forces,  R  and  S,  may  now  be  treated  in  the 
same  way  as  P  and  Q.  The  resultant  of  three  parallel  forces  which 
are  not  all  in  the  same  plane  is  equal  to  the  algebraic  sum  of  the 
forces  and  the  moment  of  the  resultant  about  any  axis  is  equal  to  the 
sum  of  the  moments  of  the  three  forces  about  that  axis.  This  method 
may  be  extended  to  include  any  number  of  forces.  The  axis  of 
moments  is  generally  in  a  plane  perpendicular  to  the  direction 
of  the  forces.  This,  however,  is  not  necessary. 

Example 

Find  the  magnitude  and  position  of  the  resultant  of  the  following  vertical 
forces:  24  pounds  down  through  the  point  (3,  5),  16  pounds  down  through 
the  point  (2,  7),  9  pounds  up  through  the  point  (5,  4),  12  pounds  down 
through  the  point  (  —  6,  3),  and  23  pounds  up  through  the  point  (8,  —5). 

In  the  tabulated  solution  below,  since  most  of  the  forces  are  downward, 
that  direction  is  taken  as  positive.  The  moments  are  calculated  about  the 
X  and  the  Z  axes. 

FORCE  x  Px  z  Pz 

24  3  72  5  120 

16  2  32  7  112 

-  9  5  -  45  4  -  36 

12  -6  -  72  3  36 

-23  8  -184  -5  115 


R  =  20  Rx  =  -197  Rz  =  347 

x  =  -9.85,  z  =  17.35, 
in  which  x  and  z  are  the  coordinates  of  the  resultant  force. 

Problems 

1.  Find  the  magnitude  and  position  of  the  resultant  of  the  following 
horizontal  forces:  184  pounds  toward  the  east  through  a  point  5  feet  above 
a  floor  and  3  feet  north  of  a  vertical  wall,  160  pounds  east  through  a  point 
8  feet  above  the  floor  and  2  feet  south  of  the  wall,  124  pounds  west  through 
a  point  3  feet  above  the  floor  and  6  feet  north  of  the  vertical  wall. 

2.  A  rectangular  door  in  a  horizontal  position  is  7  feet  long  from  left  to 
right  and  4  feet  wide.     It  weighs  40  pounds  and  its  center  of  mass  is  at  the 
center.     The  door  carries  50  pounds  at  the  right  corner  near  the  observer, 
35  pounds  at  the  left  corner  diagonally  across  from  the  50  pounds,  24  pounds 
at  the  other  right  corner,  and  16  pounds  at  the  middle  of  the  left  edge. 
Find  the  location  of  a  single  support  which  will  hold  the  door  in  the  hori- 
zontal position. 

95.  Equilibrium  of  Parallel  Forces. — To  find  the  resultant  of 
parallel  forces,  three  quantities  are  calculated.     These  are  the 


CHAP.  X]  PARALLEL  FORCES  151 

magnitude  of  the  resultant  and  two  coordinates  of  its  position. 
A  problem  of  the  equilibrium  of  parallel  forces  involves  three 
unknowns  and  requires  three  independent  equations  for  its 
solution.  Since  the  forces  are  all  in  one  direction,  each  force  is 
fully  determined  by  a  single  resolution.  Only  one  of  the  three 
possible  equations  may  be  a  resolution  equation.  When  there 
are  two  moment  equations,  one  of  the  axes  must  not  be  parallel 
to  the  other.  When  three  moment  equations,  are  written,  not 
more  than  two  of  the  axes  may  be  parallel.  It  will  be  noticed 
that  two  moment  equations  and  one  resolution  equation  were 
used  in  the  example  of  Art.  94.  It  is  not  necessary  that  the  axes 
of  moments  be  at  right  angles  to  each  other. 

Problems 

1.  A  horizontal  trap  door  is  8  feet  long  and  4  feet  wide.     Its  center  of 
mass  is  at  the  center.     The  door  has  2  hinges,  each  one  foot  from  a  corner 
on  an  8-foot  side.     The  door  is  lifted  by  a  vertical  rope  attached  3  feet  from 
one  corner  on  the  other  long  side.     Find  the  tension  in  the  rope  and  the 
hinge  reactions.     Solve  by  three  moment  equations  and  check  by  a  vertical 
resolution. 

2.  Solve  Problem  1  if  the  rope  is  attached  to  one  corner. 

3.  In  the  trap  door  of  Problem  1,  find  the  position  at  which  the  rope  may 
be  attached  at  the  edge  of  the  door  in  order  that  one  hinge  reaction  may  be 
zero.     Solve  algebraically  and  graphically. 

4.  A  horizontal  table  is  5  feet  long  from  left  to  right  and  3  feet  wide. 
It  is  supported  on  three  legs,  one  at  each  of  the  left  corners  and  one  at  the 
middle  of  the  right  side.     A  load  of  60  pounds  is  placed  1  foot  from  the 
front  edge  and  1  foot  from  the  left  edge.     Find  the  reaction  of  each  leg  if 
the  weight  of  the  table  is  neglected. 

5.  The  load  on  the  table  of  Problem  4  is  placed  6  inches  from  the  front 
edge.     How  far  may  it  be  moved  from  the  left  edge  without  upsetting  the 
table. 

6.  Solve  Problem  5  if  the  table  weighs  20  pounds  and  its  center  of  mass  is 
at  the  center. 

7.  A  wheelbarrow  is  5  feet  long  from  axis  of  axle  to  points  at  which 
the  handles  are  gripped,  and  is  2  feet  wide  between  centers  of  handles.     It 
carries  a  load  of  160  pounds,  which  is  placed  1  foot  from  the  axis  of  the  axle 
and  4  inches  to  the  right  of  the  middle.     Find  the  reaction  of  the  wheel  and 
of  each  handle. 

96.  Center  of  Gravity. — When  parallel  forces  are  applied  to 
each  particle  of  a  body,  and  each  force  is  proportional  to  the 
mass  of  the  particle  upon  which  it  acts,  the  line  of  action  of 
the  resultant  of  these  forces  passes  through  a  point  which  is 
called  the  center  of  mass  of  the  body.  The  force  of  gravity  on 


152  MECHANICS  [ART.  96 

each  particle  of  a  body  is  directed  toward  the*  center  of  the  Earth, 
which  is  so  distant  that  the  forces  on  the  various  particles  of  an 
ordinary  body  are  practically  parallel.  The  force  of  gravity 
varies  as  the  mass  of  the  particle  and  inversely  as  the  square  of 
its  distance  from  the  center  of  the  Earth.  This  distance  is  so 
great  that  for  bodies  of  ordinary  dimensions,  the  error  in  the 
assumption  that  the  attraction  on  each  particle  is  proportional 
its  mass  is  negligible.  For  these  reasons  center  of  mass  is  usually 
regarded  as  synonymous  with  center  of  gravity.  Center  of  gravity 
may  be  defined  as  the  point  of  application  of  the  resultant  of  the 
gravity  forces  which  act  on  the  body.  The  term  center  of  gravity 
is  used  more  commonly  than  center  of  mass. 

The  position  of  a  line  through  the  center  of  gravity  may  be 
found  by  taking  moments  about  two  axes  which  are  not  parallel. 
The  position  of  the  center  of  gravity  on  this  line  is  found  by 
regarding  the  direction  of  all  the  forces  as  changed  and  then 
taking  moments  about  a  third  axis. 

The  coordinates  of  the  center  of  gravity  are  represented  by 
x,  y,  and  z.  These  may  be  read  "bar  x, "  "bar  y} "  and  "bar  z." 

Example 

The  base  ABC  of  Fig.  136  is  a  plane  surface  and  the  faces  A  DDE  and 
DEC  are  also  plane  and  are  perpendicular  to  each  other  and  to  the  base. 
The  body  weighs  24  pounds.  The  body  rests  on  three  supports,  which  are 
not  shown  in  the  drawing.  Two  supports  are  1  inch  from  the  edge  AB. 

The  third  support,  17  inches  from  the 
edge  AB,  rests  on  a  platform  scale. 
The  scale  reading  is  4.52  pounds 
when  the  base  is  horizontal.  How 
far  is  the  center  of  gravity  from  the 
plane  ABDE? 

Taking  moments   about  the  two 
supports, 

FIG.  136.  24x  =  4.52  X  16, 

x  =  3.01  in. 

The  center  of  gravity  lies  in  a  vertical  plane  which  is  parallel  to  the  face 
ABDE  and  is  4.01  inches  from  it. 

When  the  body  is  supported  on  two  points,  each  1  inch  from  the  edge  BC, 
and  a  third  point  rests  on  a  platform  scale  13  inches  from  BC,  the  scale 
reading  is  found  to  be  7.24  pounds.  From  this  experiment,  the  center  of 
gravity  is  found  to  lie  in  a  plane  parallel  to  the  face  BCD  at  a  distance  of 
4.61  inches  from  that  plane.  The  intersection  of  these  two  planes  through 
the  center  of  gravity  gives  the  location  of  a  line  through  the  center  of  gravity. 

To  find  the  location  of  the  center  of  gravity  on  the  line  of  intersection 


CHAP.  X]  PARALLEL  FORCES  153 

of  the  two  planes,  its  position  must  be  changed  so  that  the  forces  of  gravity 
will  act  in  a  relatively  different  direction.  For  the  experiment  the  body  is 
placed  with  the  face  BCD  horizontal.  When  it  is  supported  at  two  points, 
each  1  inch  from  BC,  and  on  a  third  point  at  a  distance  of  11  inches  from 
BC,  the  scale  supporting  the  third  point  reads  7.20  pounds.  From  this 
reading  the  center  of  gravity  is  found  to  be  4.00  inches  from  the  plane 
ABC. 

It  will  be  observed  that  each  of  these  experiments  locates  a  vertical  plane 
which  is  parallel  to  the  axis  of  moments  and  passes  through  the  center  of 
gravity.  Three  such  planes,  if  no  two  of  them  are  parallel,  fully  locate  the 
center  of  gravity. 

Problems 

1.  A  triangular  board  of  uniform  thickness  is  12  inches  wide  at  the  base. 
The  perpendicular  distance  from  the  vertex  to  the  base  is  18  inches.     The 
board  is  supported  on  two  knife-edges,  each  0.5  inch  from  the  base,  and  on  a 
third  knife  edge,  parallel  to  the  base,  at  1  inch  from  the  vertex.     The  board 
weighs  15.00  pounds.     When  the  third  knife-edge,  which  weighs  0.45  pound, 
rests  on  a  platform  scale,  the  reading  is  5.45  pounds.     How  far  is  the  center 
of  gravity  of  the  board  from  the  base?  Ans.     6.00  inches. 

2.  A  rectangular  plate  of  uniform  thickness  is  24  inches  long  and  18 
inches  wide.     A  circular  hole,  10  inches  in  diameter,  is  cut  from  the  plate. 
The  center  of  the  hole  is  6  inches  from  the  front  24-inch  edge  and  8  inches 
from   the  left  18-inch  edge.     The  plate  is  supported  on  a  knife-edge  0.1 
inch  from  the  left  edge  and  on  a  second  knife  edge  at  the  same  distance 
from  the  right  edge.     The  knife  edge  at  the  right  rests  on  a  platform  scale. 
The  increase  of  the  load  on  the  scale  when  the  plate  is  put  on  the  knife-edge, 
amounts  to  22.72  pounds.     The  plate  weighs  42.40  pounds.     How  far  is 
its  center  of  gravity  from  the  left  edge? 

As  a  second  experiment,  the  plate  is  supported  on  a  knife-edge  0.1  inch 
from  the  front  and  on  a  second  knife-edge  17  inches  from  this  one.  The 
load  on  the  second  knife-edge  reads  23.87  pounds.  How  far  is  the  center 
of  gravity  from  the  front  edge  of  the  plate? 

3.  A  heavy  rectangular  door  is  placed  in  a  horizontal  position  and  sup- 
ported by  three  spherical  steel  balls.     Each  ball  rests  on  a  separate  platform 
scale.     The  door  is  6  feet  long  from  left  to  right  and  4  feet  wide.     One  ball 
is  1  inch  from  the  front  edge  and  1  inch  from  the  left  edge.     The  scale  which 
supports  it  reads  32  pounds.     The  second  ball  is  1  inch  from  the  left  edge 
and  3  inches  from  the  rear  edge.     The  scale  which  supports  it  reads  104 
pounds.     The  third  ball  is  1  inch  from  the  right  edge  and  1  foot  from  the 
front  edge.     The  scale  which  supports  it  reads  144  pounds.     Locate  the 
center  of  gravity. 

Ans.     One  inch  to  the  right  and  1  inch  to  the  front  of  the  middle. 

When  the  dimensions  of  a  body  and  its  density  at  all  points 
are  known,  the  center  of  gravity  may  be  calculated.  Moments 
are  taken  in  the  same  way  as  in  the  experimental  determination. 
For  the  last  determination,  the  position  of  the  body  is  not  changed. 


154  MECHANICS  [ART.  97 

The  direction  of  the  parallel  forces  is  assumed  to  be  different. 
This  assumption  makes  it  possible  to  take  moments  about  an 
axis  which  is  perpendicular  to  the  first  two  axes  of  moments. 
In  Fig.  136,  the  forces  may  be  regarded  as  parallel  to  the  line 
AB. 

When  all  the  forces  are  parallel  to  a  given  plane,  the  moment 
about  any  one  of  a  set  of  parallel  lines  in  that  plane  is  the  same. 
In  this  case  the  moment  is  often  called  the  moment  with  respect 
to  the  plane.  In  Fig.  136,  the  moment  about  AB  is  equal  to 
the  moment  about  any  line  in  the  plane  of  ABDE  parallel  to 
the  line  AB.  The  moment  with  respect  to  AB  is  the  moment 
with  respect  to  the  plane  ABDE. 

97.  Center  of  Gravity  Geometrically. — When  the  density  of  a 
body  is  uniform,  the  center  of  gravity  may  sometimes  be  deter- 
mined geometrically.  If  the  body  has  a  plane  of  symmetry,  the 
plane  divides  it  into  equal  halves.  If  the  plane  of  symmetry 
be  placed  vertical,  or  if  the  forces  be  regarded  as  acting  parallel 
to  it,  the  moment  of  one-half  of  the  body  about  any  axis  in  the 
plane  is  equal  and  opposite  the  moment  of  the  other  half.  //  a 
body  of  uniform  density  has  a  plane  of  symmetry,  the  center  of 
gravity  lies  in  that  plane.  If  there  are  two  or  more  planes  of 
symmetry,  the  center  of  gravity  lies  at  their  intersection. 

All  planes  through  the  center  of  a  sphere  are  planes  of  sym- 
metry. The  center  of  gravity  of  a  sphere  is  at  its  center. 

All  planes  through  the  axis  of  a  right  circular  cylinder  are 
planes  of  symmetry.  The  plane  perpendicular  to  the  axis  at 
the  middle  is  also  a  plane  of  symmetry.  The  center  of  gravity 
of  a  right  circular  cylinder  of  uniform  density  is  located  at  the 
middle  of  the  axis. 

The  center  of  gravity  of  a  rectangular  parallelopiped  is  located  at  a 
point  which  is  midway  between  all  the  faces. 

A  The  center  of  gravity  of  a  right  prism,  the  sec- 

tions of  which  are  equilateral  triangles  is  located 
at  the  middle  of  its  length  at  the  intersection  of 
the  lines  which  bisect  the  angles  of  the  section. 

Figure  137  shows  an  oblique-angled  parallelo- 
gram. The  sections  perpendicular  to  the  plane 
of  the  paper  are  assumed  to  be  rectangular.  The 
plane  of  which  the  line  AB  is  the  trace  is  not 
a  plane  of  symmetry.  It  is  a  plane  of  symmetry  for  elements, 
such  as  EFG,  perpendicular  to  it,  but  these  elements  do  not 


CHAP.   X] 


PARALLEL  FORCES 


155 


include  the  entire  figure.  They  leave  out  a  triangular  area  at 
the  top  and  another  at  the  bottom.  If,  however,  the  parallelo- 
gram is  divided  into  elements  such  as  COD,  by  lines  parallel  to 
the  ends,  a  series  of  such  elements  will  include  the  entire  figure. 
It  will  be  shown  that  the  center  of  gravity  of  the  element  COD 
is  midway  between  the  ends.  If  the  part  OC  is  rotated  about  the 
line  AB  as  an  axis,  it  will  not  coincide  with  the  part  OD.  If, 
however,  OC  is  rotated  through  180  degrees  about  an  axis  at  0 
perpendicular  to  the  plane  of  the  paper,  it  will  coincide  exactly 
with  OD.  The  center  of  gravity  of  OC  is  as  far  from  the  line  AB 
on  one  side  as  the  center  of  gravity  of  OD  from  the  line  on  the 
other  side.  The  moment  of  OC  about  the  line  AB  is  equal  and 
opposite  to  the  moment  of  OD.  The  center  of  gravity  of  the 
element  lies  in  the  plane  of  AB.  The  center  of  gravity  of  an 
oblique  prism  or  cylinder  lies  in  a  plane  parallel  to  the  bases  and 
midway  between  them. 

Problems 


1.  A  plate  of  uniform  thickness  and  density  is  of  the  form  of  a  5' 
by  1"  angle  section  as  shown  in  Fig.  138.  Find 
the  distance  of  the  center  of  gravity  from  the 
back  of  each  leg.  Solve  by  dividing  the  section 
into  two  equal  rectangles.  Check  x  by  dividing 
the  section  into  a  5"  by  1"  rectangle  and  a 
1"  by  3"  rectangle.  Since  the  thickness  and 
density  are  constant,  the  terms  which  represent 
them  may  be  omitted. 

Taking  moments  about  the  back  of  the  4-inch 
leg, 

AREA  ARM  MOMENT 

4  0.5  2.0 

4  3.0  12.0 


by  4" 


8 


14.0 


Total  mass 


=  8.     Total  moment 
8z  =  14, 
*  =  1.75  in. 


14 


2.  A  homogeneous  solid  cylinder,  6  inches  in  diameter  and  10  inches  long, 
weighs  15  pounds.     The  cylinder  stands  on  a  rectangular  parallelepiped,  10 
inches  square  and  6  inches  high,  which  weighs  35  pounds.     Find  the  distance 
of  the  center  of  gravity  of  the  two  from  the  base  of  the  parallelepiped. 

Ans.     5Qy  =  270;  y  =  5.4  in.  from  the  base. 

3.  A  homogeneous  solid  cylinder  8  inches  in  diameter  is  6  inches  long.     A 
second  homogeneous  solid  cylinder  is  4  inches  in  diameter  and  10  inches 
long.     The  8-inch  cylinder  stands  vertical  and  the  4-inch  cylinder  is  placed 


156  MECHANICS  [ART.  98 

on  it  with  its  axis  2  inches  to  the  right  of  the  axis  of  the  first  cylinder.     If 
both  cylinders  have  the  same  density,  find  the  center  of  gravity. 

Ans.     x  =  0.59  in.  from  axis  of  8-in.  cylinder; 
y  =  5.35  in.  from  base  of  8-in.  cylinder. 

4.  Solve  Problem  2  if  the  cylinder  and  block  have  the  same  density. 
6.  Solve  Problem  3  if  the  density  of  the  8-inch  cylinder  is  twice  that  of  the 
4-inch  cylinder. 

6.  The  plate  of  Problem  2  of  Art.  96  is  of  uniform  thickness  and  density. 
Determine  the  position  of  the  center  of  gravity.     Subtract  the  moment  and 
mass  of  the  circular  hole  from  the  moment  and  mass  of  the  entire  plate. 

7.  A  homogeneous  cylinder  8  inches  in  diameter  and  10  inches  long  has  a 
circular  hole  drilled  in  one  end.     The  hole  is  4  inches  in  diameter,  8  inches 
deep,  and  its  axis  is  1  inch  from  the  axis  of  the  cylinder.     Find  the  center  of 
gravity  of  the  remainder. 

98.  Center  of  Gravity  of  a  Triangular  Plate.— Figure  139  repre- 
sents a  triangular  plate  of  uniform  thickness  perpendicular  to  the 


H 
FIG.  139.  FIG.  140. 

plane  of  the  paper.  OH  is  the  median  line.  The  triangle  is 
divided  into  elements  such  as  ADFB,  by  lines  parallel  to  the 
base.  The  moment  of  the  oblique  parallelogram  AC  ED  with 
respect  to  the  line  0  H  (or  any  line  in  the  plane  through  OH 
perpendicular  to  the  plane  of  the  paper)  is  zero.  The  small 
triangles  ABC  and  DEF  are  left  over.  These  triangles  have  equal 
areas,  and  their  bases  are  equally  distant  from  the  line  OH',  con- 
sequently their  moments  about  OH  must  be  approximately  equal. 
The  lines  A  D  and  B  F  may  be  brought  infinitely  close  together, 
so  that  the  areas  of  these  triangles  become  infinitesimals  of  the 
second  order,  and  the  difference  of  their  moments  become  infini- 
tesimals of  the  third  order.  The  center  of  gravity  of  the  ele- 
ment ADFB  must  then  lie  in  the  median  line.  The  entire 
triangle  is  made  up  of  similar  elements,  each  of  which  has  its 
center  of  gravity  on  the  median  line.  It  follows  that  the  center 
of  gravity  of  the  entire  triangular  plate  of  uniform  thickness  and 
density  must  lie  on  the  plane  of  the  median  line. 


CHAP.  X]  PARALLEL  FORCES  157 

A  second  median  may  be  drawn  as  in  Fig.  140.  The  center 
of  gravity  lies  in  this  median,  and,  consequently,  at  the  intersec- 
tion of  the  medians. 

It  was  learned  in  Geometry  that  the  intersection  of  the  medians 
of  a  triangle  is  located  at  one-third  the  length  from  the  base.  In 
Fig.  140,  HK  is  one-third  of  OH,  and  KL  is  one-third  of  ML.  If 
h  is  the  perpendicular  distance  from  a  vertex  to  the  opposite  base, 
and  y  is  the  perpendicular  distance  from  the  center  of  gravity 
to  that  base,  then  y  is  one-third  of  h. 

Example 

A  plate  of  uniform  thickness  and  density  is  in  the  form  of  a  trapezoid 
The  lower  base  is  18  inches,  the 
upper  base  is  10  inches,  and  the 
height  is  12  inches.  Find  the 
distance  of  the  center  of  gravity 
from  the  lower  base. 

The  trapezoid  may  be  divided, 
as  shown  in  Fig.  141,  I,  into  a 
parallelogram  and  a  triangle. 
Since  the  thickness  and  density 

are  constant,  their  values  in  the  mass  and  moment  will  cancel.     Only  the 
area,  then,  need  be  considered. 

AREA  ARM  MOMENT 

Parallelogram 120  6  720 

Triangle 48  4  192 

IQSy  912 

y  =  5f  in. 


Problems 


1.  Solve  the  example  above  by  dividing  the  trapezoid  into  two  triangles 
as  in  Fig.  141,  II. 

2.  Solve  the  example  above  by  means  of  a  parallelogram  of  18-inch  base 
from  which  an  inverted  triangle  is  subtracted  to  form  the  trapezoid. 

3.  A  trapezoidal  plate  of  uniform 
thickness  has  the  lower  base  24  inches, 
the  upper  base  18  inches,  and  the 
height  15  inches.  Find  the  distance  of 
the  center  of  gravity  from  the  lower 
FIG.  142.  base.  Check  by  moments  about  the 

upper  base. 

4.  Figure  142  shows  a  standard  15"  channel  section.  Find  the  distance 
of  the  center  of  gravity  from  the  back  of  the  web,  and  compare  the  result 
with  Cambria  Steel  or  Carnegie  Handbook. 


158  MECHANICS  [ART.  99 

6.  A  wooden  block  is  18  inches  square  and  10  inches  high,  and  its  specific 
gravity  is  0.8.  A  metal  cylinder,  5  inches  in  diameter  and  4  inches  high, 
stands  on  the  block.  The  axis  of  the  cylinder  is  10  inches  from  the  front 
face  of  the  block  and  12  inches  from  the  left  face.  The  specific  gravity 
of  the  metal  is  7.8.  Locate  the  center  of  gravity  of  both  with  reference  to 
the  bottom,  left,  and  front  surfaces  of  the  block. 

Ans.     y  =  6.34  in.;  x  =  9.57  in.;  z  =  9.19  in. 

6.  A  triangular  steel  plate  is  2  inches  thick.  The  triangle  is  isosceles 
with  a  base  of  18  inches  and  an  altitude  of  30  inches.  Steel  weighs  500 
pounds  per  cubic  foot.  The  plate  is  placed  horizontally  and  supported  by 
three  spherical  balls.  One  ball  is  1  inch  from  the  base  and  6  inches  from 
the  bisector  of  the  opposite  angle.  Another  sphere  is  1  inch  from  the  base 
and  8  inches  from  the  bisector.  The  third  sphere  is  on  the  bisector  at  a 
distance  of  24  inches  from  the  base.  Find  the  load  on  each  sphere. 


99.  Center  of  Gravity  of  a  Pyramid. — Figure  143  shows  a  tri- 
angular pyramid.  A  median  BE  is  drawn  in  the  base  BCD,  and 
a  point  F  is  located  at  one-third  the  length  of  the  median  from 
E.  If  the  base  is  regarded  as  a  thin  triangular  plate,  the  point 
F  is  at  its  center  of  gravity.  A  line  AF  is  drawn  from  the  vertex 
A  to  the  center  of  gravity  of  the  base.  Let  B'C'D'  be  a  thin 
plate  between  two  planes  which  are  parallel  to  the  base  BCD. 
This  plate  is  similar  to  the  base.  From 
the  similarity  of  the  triangles  it  may  be 
shown  that  the  line  AF  passes  through 
the  center  of  gravity  of  this  plate.  Since 
B'C'D'may  be  any  plate  parallel  to  the 
base,  it  follows  that  the  center  of  gravity 
of  the  pyramid  is  located  on  the  line  AF. 
In  the  face  ACD  the  median  AE  is 
drawn  and  the  point  G  is  located  on  this 
median  at  one-third  of  its  length  from 

E.  The  line  BG,  from  the  vertex  opposite  the  face  ACD,  passes 
through  the  center  of  gravity  of  the  pyramid.  The  lines  AF  and 
BG  are  in  the  plane  AEB  and  intersect  at  the  point  H  in  that 
plane. 

The  triangles  ABE  and  GFE  have  the  same  angle  at  E.  The 
side  GE  is  one-third  of  AE  and  the  side  FE  is  one-third  of  BE. 
The  triangles  are  similar  and  the  side  GF  is  parallel  to  A  B  and 
one-third  its  length.  In  the  triangles  BAH  and  GFH,  the  vertical 
angles  at  H  are  equal.  The  angle  at  G  is  equal  to  the  angle 


CHAP.  X]  PARALLEL  FORCES  159 

at  B  and  the  angle  at  E  is  equal  to  the  angle  at  A.     These 
triangles  are  similar  and 

HF  =  GF      i 
AH      AB      3; 


4  HF  =  AF;  AF  =    ~ 

The  center  of  gravity  of  a  triangular  pyramid  is  on  a  line  joining 
the  vertex  with  the  intersection  of  the  medians  of  the  base,  at  a  distance 
from  the  base  equal  to  one-fourth  the  length  of  this  line.  If  h  is  the 
vertical  distance  of  the  vertex  from  the  base,  the  center  of  gravity 

lines  in  a  plane  parallel  to  the  base  at  a  distance  of  j  therefrom. 

A  pyramid  with  quadrilateral  base  may  be  divided  into  two 
triangular  pyramids.  Since  the  center  of  gravity  of  each  of  these 
triangular  pyramids  is  at  a  distance  of  one-fourth  the  vertical 
height  from  the  base,  the  center  of  gravity  of  the  entire  pyramid 
lies  in  the  same  plane.  This  may  be  extended  to  pyramids  of 
any  form. 

A  cone  may  be  regarded  as  made  up  of  an  infinite  number  of 
small  pyramids  of  the  same  altitude.  Its  center  of  gravity  is, 
therefore,  at  one-fourth  the  height  from  the  base. 

Problems 

1.  A  homogeneous  solid  right  cylinder,  5  inches  in  diameter  and  6  inches 
long,  stands  with  its  axis  vertical.     On  the  top  of  the  cylinder  is  placed  a 
homogeneous  solid  right  cone,  which  is  6  inches  in  diameter  and  8  inches 
high.     The  axis  of  the  cone  is  1  inch  from  the  axis  of  the  cylinder.     The 
density  of  cone  and  cylinder  is  the  same.     Find  the  center  of  gravity  of 
the  combination.  AnSi     x  =  0.39  in.  from  the  axis  of  cylinder; 

y  =  4.95  in.  from  the  base  of  cylinder. 

2.  A  homogeneous  solid  right  cylinder  is  10  inches  in  diameter  and  8 
inches  high.     A  cone  6  inches  in  diameter  and  6  inches  deep  is  cut  out  of  the 
top  of  the  cylinder.     The  axis  of  the  cone  is  1  inch  from  the  axis  of  the 
cylinder.     Find  the  center  of  gravity  of  the  remainder. 

3.  Solve  Problem  1  if  the  density  of  the  cylinder  is  twice  the  density  of 
the  cone. 

4.  A  frustrum  of  a  cone  is  6  inches  in  diameter  at  the  lower  base,  4  inches 
in  diameter  at  the  upper  base,  and  6  inches  high.     Find  its  center  of  gravity. 

100.  Center  of  Gravity  by  Integration. — The  calculation  of 
the  center  of  gravity  is  equivalent  to  the  location  of  the  line 


160  MECHANICS  [ART.  100 

of  application  of  the  resultant  of  parallel  forces.  The  moment  of 
each  mass  is  found  with  respect  to  some  plane;  then  this  moment 
is  divided  by  the  sum  of  the  masses.  If  m\  is  the  mass  of  one 
element  of  the  body,  and  x\  is  its  abscissa,  its  moment  about  any 
line  in  the  YZ  plane  is  m\x\.  If  mi,  m2,  Ws,  etc.  are  masses,  and 
Xit  #2,  £3,  etc.,  are  the  coordinates  of  their  centers  of  gravity  with 
respect  to  some  plane  of  reference,  the  total  moment  with  respect 
to  that  plane  is  ' 


M  =  m\x\  -f-  m^Xz  +  mzx3  +  etc.  =  Smx.  (1) 

Total  mass  =  Wi+  m%  +  ^3  +  etc.  =  2m.  (2) 


Similarly,  the  other  coordinates  of  the  center  of  gravity  are 

2m?/ 
y  =  -- 


2mz 


In  order  to  use  these  equations,  the  center  of  gravity  of  each 
of  the  parts  which  compose  the  system  must  be  known.  When 
the  center  of  gravity  of  each  part  is  not  known,  it  is  necessary 
to  divide  the  body  into  infinitesimal  elements,  write  the  expres- 
sion for  the  moment  of  each  element,  and  find  the  sum  of  the 
moments  by  integration.  If  dm  is  the  mass  of  an  element  and  x' 
is  the  coordinate  of  the  center  of  gravity,  Equation  (1)  becomes 

M=  fx'dm.  (6) 

If  p  is  the  density  and  dV  is  the  element  of  volume, 

dm  —  pdV, 
M  =  fpx'dV.  (7) 

If  p  is  the  mass  of  a  plate  per  unit  area,  and  dA  is  an  element  of 
area, 

dm  =  pdA, 
M  =  fpx'dA. 

If  the  density  is  uniform  throughout  all  parts  of  the  body,  it 
may  be  omitted  in  finding  the  center  of  gravity. 
With  the  integral  expressions  for  the  moment, 
fx'dm_  fpx'dV 
fdm          fpdV  '  W 


CHAP.  X] 


PARALLEL  FORCES 


When  the  density  is  constant,  Equation  (8)  becomes 
_  fx'dV       fx'dV 

~7W      ~F~ 

For  a  plate  of  uniform  thickness  and  density, 


161 


Similar  expressions  give  y  and  z. 

It  must  be  remembered  that  xr  is  the  distance  of  the  center  of 
gravity  of  the  element  from  the  plane  of  reference. 

Example  I 

Find  the  center  of  gravity  of  a  triangular  plate  of  uniform  thickness  and 
density,  by  moments  about  an  axis  through  the 
vertex  parallel  to  the  base.     (Fig.  144.) 

The  element  of  area  is  y  high  and  dx  wide. 
This  form  of  element  is  chosen  because  the  entire 
triangle  may  be  built  up  of  strips  of  this  kind  and 
because  its  moment  arm  is  easy  to  find.  Neglect- 
ing the  small  triangle  at  the  top  of  the  element, 
its  area  is  ydx.  The  moment  arm  of  the  ele- 
ment about  the  line  AB  through  the  vertex  of  the 
triangle  is 


FIG.  144. 


When  -~-  is  multiplied  by  the  element  of  area,  the  product  is  a  differential 
of  the  second  order  and  may  be  neglected. 

M  =  fx'dA  =  fxydx.  (11) 

This  expression  for  the  moment  contains  two  variables,  x  and  y.  One  of 
these  variables  must  be  eliminated.  From  the  geometry  of  the  similar 
triangles  of  Fig.  144, 


bx 


Substituting  in  Equation  (11)  and  integrating, 

_  Cbx2  d         b    C    „ ,         b 

The  limits  for  x  are  x  =  0  and  x  =  h, 


o~    3 


Dividing  the  moment  by  the  area, 


bh*      bh      2h 


11 


162 


MECHANICS 


[ART.   100 


Problem 

1.  Instead  of  eliminating  y  from  Equation  (11)  of  Example  I,  eliminate  x 
and  dx  and  integrate  between  the  proper  limits  for  the  moment.  Also 
integrate  ydx  for  the  area  of  the  triangle. 

Example  II 

Find  the  center  of  gravity  of  a  triangle  by 
means  of  elements  perpendicular  to  the  base. 

In  Fig.  .145;  the  vertical  line  CD  is  taken 
as  the  base  of  the  triangle.  With  the  origin 
at  the  vertex,  x  is  the  abscissa  of  a  point  on 
the  line  OC.  The  length  of  an  element  of  area 
is  h  —  x  and  the  width  is  dy 

FIG.  145.  .         *  dA  =  (h  -  x)dy. 

The  center  of  gravity  of  the  element  is  at  the  middle  of  its  length,  which  is 
at  a  distance  of  — ~ —  from  either  end. 


h  -  x      h  +  x 


M 


dy. 


(13) 


(14) 


Since  y  =  -j->  dy  =  T  dx.     Substituting  in  Equation  (14) 


M 


-  x')dx  -    -     h'x-  -  — 

~ 


3 


This  result  applies  to  the  right  triangle  OCE.  Dividing  by  the  area,  the 
center  of  gravity  of  this  right  triangle  is  found  to  be  at  two-thirds  the  altitude 
from  the  vertex.  In  a  similar  way  the  center  of  gravity  of  the  other  right 
triangle  OED  is  found  to  be  at  the  same  distance  from  the  vertex.  Since 
the  center  of  gravity  of  the  entire  triangle  is  on  the  line  which  joins  these  two 
centers,  it  is  located  at  two-thirds  the  altitude  from  the  vertex. 


Problems 

2.  Substitute  for  x2  in  Equation  (14)  and  integrate  in  terms  of  x  as  the 
independent  variable. 

3.  Find  the  center  of  gravity  of  a  quadrant  of  a  uniform  circular  plate 
of  radius  a. 

Using  the  vertical  element  of  Fig.  146  to  find  the  moment  with  respect 
to  the  plane  of  0 A , 


3 


a 

*     ~~     O 

3 


4a 

0~"' 

3ir 


Solve  for  y  by  means  of  the  same  element. 


CHAP.   X] 


PARALLEL  FORCES 


163 


4.  Find  the  center  of  gravity  of  a  60-degree  sector  of  a  circle  of  radius  a 
by  polar  coordinates  with  double  integration. 

The  element  of  area  is  rdedr,  and  the  moment  arm  for  finding  x  is  x'  = 
r  cos  0 


FIG.  146. 


FIG.  147. 


M 
M 


oV3 
6 

~6~ 


cos  e  de  dr  =  —  /cos  6  de  =  — 
o  o 


^-  ?ra2  =  0/3 
6  TT  ' 


The  Y  coordinate -of  the  center  of  gravity  may  be  found  by  a  similar 
method,    and   may  oe   checked  by  the  condition 
that  the  center  of  gravity  lies  on  the  bisector  of 
the  angle. 

6.  Find  the  center  of  gravity  of  a  60-degree 
sector  of  a  circle  of  radius  a,  using  the  bisector  of 
the  angle  as  the  axis  of  X.  Check  trigonom- 
etrically  with  the  results  of  Problem  4. 

6.  Find  the  center  of  gravity  of  a  wire  of 
uniform  section  which  is  bent  into  the  form  of 
a  semicircle  of  radius  a.  Use  polar  coordinates 
with  a  single  integraion.  » 

Ans.     x  —  — :  y  =  0. 


FIG  148. 


7.  Find  the  center  of  gravity  of  a  homogeneous  solid  hemisphere, 
the  hemisphere  of  circular  disks  as  in  Fig.  148. 

Ans.     M  =  7r/(a2  - 


Build 


7T04 


3a 


8.  A  homogeneous  solid  cylinder,  8  inches  in  diameter  and  6  inches  long 
has  a  hemispherical  depression,  6  inches  in  diameter,  in  one  end.  Find 
the  center  of  gravity  of  the  remainder. 


164 


MECHANICS 


[ART.    100 


9.  Find  the  center  of  gravity  of  any  cone  of  altitude  h.     Build  up  the 
cone  of  flat  disks  parallel  to  the  base. 

10.  Find  the  center  of  gravity  of  a  right  cone  by  integration.     Build 
up  the  cone  of  concentric  hollow  cylinders  as  in  Fig.  149. 

11.  Find  the  center  of  gravity  of  a  hemisphere  by  integration,  using 
hollow  cylinders  as  elements  of  volume. 


FIG.  149. 


Example  III 

Find  the  center  of  gravity  of  a  segment  of  a  homogeneous  sphere  of  radius 
a  cut  off  by  a  plane  at  a  distance  of  ~  from  the  center. 

In  this  case  the  volume  of  the  body  is  not  known  and  must  be  obtained 
by  integration.  Using  an  element  in  the  form  of  a  circular  disk,  as  in  Fig. 
148,  dV  =  y*dx,  and  x'  =  x. 

I~     2    2  4T  o 

rSxyUx       Tr/Ca'  -  x*)xdx       *  I  2     ~  4  Jf. 


64 


27 


24 


The  factor  TT  in  the  numerator  and  denominator  might  have  been 
canceled  at  the  beginning.  It  is  carried  through,  however,  in 
order  to  have  the  true  moment  and  volume  at  the  end. 

Do  not  make  the  mistake  of  canceling  terms  to  the  right  of  the 
integral  signs.  After  integrating,  common  factors  in  the  numera- 
tor and  denominator  may  be  canceled  before  the  limits  are  put 
in,  provided  one  limit  is  zero.  It  is  safer,  however,  not  to  cancel 
any  terms  until  after  the  limits  are  put  in. 

Problems 

12.  Make  an  approximate  check  of  Example  III,  by  considering  a  seg- 
ment to  be  replaced  by  a  cone. 


CHAP.   X] 


PARALLEL  FORCES 


165 


13.  Find  the  center  of  gravity  of  a  uniform  plate  in  the  form  of  a  segment 
of  a  circle  of  radius  a  which  is  cut  off  by  a  plane  at  a  distance  of  §  from  the 

o 

center.     Use  a  vertical  element  as  in  Fig.  150.     When  integrating  for  the 
area,  substitute  x  =  a  cos  6,  y  =  a  sin  6. 

Ans.     x  =  O.GlOa:  y  =  0. 


FIG.  150. 


FIG.  151. 


14.  Find  y  for  half  the  segment  of  Problem  13.     Use  the  area  already 
obtained  for  the  entire  segment  and  integrate  for  the  moment  only. 

15.  A  uniform  plate  is  bounded  by  the  X  axis,  the  curve  y2  =  4x  and 
the  lines  x  =  4  and  x  =  9.     Find  its  center  of  gravity,  Fig.  151. 

Ans.     x  =  6.66;  y  =  2.57. 

16.  Check  Problem  15  approximately  by  replacing  the  area  by  the  enclosed 
trapezoid. 

17.  Find  x  for  the  area  bounded  by  the  X  axis,  the  line  x  =  a,  and  the 
curve  y2  =  bx.  Ans.    x  =  0.6a. 

18.  Find  x  for  the  area  bounded  by  the  X  axis,  the  line  x  =  a,  and  the 


line  y  =  bx. 

19.  Find  x  for  the  area  bounded  by  the  X  axis,  the 
line  x  =  a,  and  the  curve  y  =  bx2. 

3a 

Ans.     x  =  -r- 

20.  Find  x  for  the  area  bounded  by  the  X  axis,  the 
line  x  =  a,  and  the  curve  y  =  bx3. 

4a 

Ans.     x  =  -=-• 
o 

21.  A   solid   cylinder  is  cut  by  two  planes  which 
intersect  on  a  diameter.    One  of  these  planes  is  per- 
pendicular to  the  cylinder  and  the  other  makes  an 
angle  a  with  this  plane,  Fig.  152.     Find  the  center  of 
gravity  of  the  part  of  the  cylinder  between  the  planes. 


Ans. 


Ans.     £  = 


16 


FIG.  152. 


22.  A  solid  cylinder  of  radius  a  is  cut  by  two  planes  which  intersect  on  a 
tangent  and  make  an  angle  a  with  each  other.  One  plane  is  normal  to  the 
axis  of  the  cylinder.  Find  the  center  of  gravity  of  the  portion  of  the 
cylinder  between  the  planes.  Solve  by  polar  coordinates. 


166 


MECHANICS 


[ART.    101 


101.  Combination  Methods  of  Calculation.  —  It  frequently 
happens  that  part  of  an  area  or  volume  may  be  calculated  by 
arithmetical  methods,  while  the  remainder  requires  the  use  of  the 
calculus. 

Example 

Solve  Problem  15  of  Art.  100  with  horizontal  elements  of  area,  Fig.  153. 

There  are  two  sets  of  elements.     The  elements  of  one  set  extend  from 

x  =  4  to  x  =  9;  the  elements  of  the  other  set,  from  the  curve  to  x  =  9. 

The  elements  of  the  first  set  form  a  rectangle  so  that  it  is  not  necessary  to 

integrate    for    the   area  and   moment. 
The  area  of  this  rectangle  is  20 


20X2+ 


'/(9-  j 

~ 


20  + 


FIG.  153. 


s  =  2.57. 


Problems 


1.  Find  the  center  of  gravity  of  the  area  bounded  by  the  X  and  Y  axes 
the  curve  xy  =  48,  the  line  x  =  12,  and  the  line  y  =  8. 

Ans.     x  =  5.32;  y  =  ? 

2.  Find  the  center  of  gravity  of  the  area  enclosed  between  the  curve 
xy  —  48  and  a  circle  of  radius  10  with  its  center  at  the  origin. 

3.  Find  the  center  of  gravity  of  the  area  enclosed  by  the  curve  y2  =  4x, 
the  line  y  =  2,  and  the  line  x  =  9. 

102.  Center  of  Gravity  of  Some 
Plane  Areas. — The  so-called  center 
of  gravity  of  a  plane  area  is  a  factor  of 
great  importance  in  the  study  of 
Strength  of  Materials.  A  plane  area 
is  equivalent  to  a  plate  of  uniform 
thickness  and  density. 

Figure  154  shows  a  plane  area  bounded  by  the  X  axis,  the 
curve  y  =  kxn,  and  the  ordinate  x  =  a.  The  length  of  the 
extreme  ordinate,  where  x  =  a,  is  y  =  kan  =  b. 

To  find  the  area  with  a  vertical  element, 

dA  =  kxndx', 

*.-y-xtf 

In  H-  Uo 
In  a  triangle,  n  =  1;  A  =  -^- 


FIG.  154. 


kan+1 


ab 


n  +  l 


(1) 


CHAP.  X] 


PARALLEL  FORCES 


167 


In  a  parabola  with  the  axis  horizontal,  n  —  ~;  A  =  -3— 

^  o 

In  a  parabola  with  axis  vertical,  n  =  2;  A  =  -~- 
To  find  the  moment  with  respect  to  the  YZ  plane, 
M  =    Ix'dA  =  k  ix»+ldx  = 


M  = 


kan+z 


= 


n  +2 


1 


In  a  triangle,  where  n  =  1,  x  =  -~- 

In  a  parabola  with  axis  horizontal,  where  n  =  s'  ^  =  ~H~* 

In  a  parabola  with  axis  vertical,  where  n  =  2,  x  =  -j- 


(2) 
(3) 


(4) 


168  MECHANICS  [ART.   102 

To  find  the  moment  with  respect  to  the  XZ  plane, 


M  = 
M  = 


Cy 


ab* 


2(2n 

M 

A        2(2n  +  1) 


1)       2(2n  +  1) 
afr2          ;       ab 


(n 


n  +  1       2(2n 
In  a  triangle,  where  n  =  I,  y  =  ^  • 
In  a  parabola  with  axis  horizontal,  where  n  =  ~ 


1) 


(5) 
(6) 
(7) 


36 


Figure  155  shows  four  of  the  most 
important  cases  of  areas  bounded  by  the 
X  axis,  the  ordinate  x  =  a,  and  the  curve 
y  =  kxn.  It  shows  also  the  center  of 
gravity  of  a  90-degree  sector  and  a  60- 
degree  sector. 

Figure  156  shows  any  sector  of  a  circle 
of  radius  a  bounded  on  one  side  by  the  X 
axis,  and  on  the  other  by  a  radius  at  an 
angle  with  the  X  axis.  The  area  is  ira?  multiplied  by  the  ratio  of 
the  arc  to  the  circumference  of  the  circle.  If  a °  be  the  angle  in 

2° 
degrees,  A     = 


FIG.  156. 


To  find  the  moment  with  respect  to  the  YZ  plane  with  polar 
coordinates, 

M  =    I  r2  cos  6  d6  dr  =  ^  I  cos  6  dO  =  ^  [sin  B]Q°.          (8) 
J  6J 

(9) 


£   = 


M 


120  a  sin  a;c 


A        3  "  360 

For  a  60-degree  sector,  sin  a  = 

120a 


and 


X    = 


607T  7T 

Since  these  sectors  are  symmetrical  with  respect  to  the  radius 
at  angle  |  with  the  X  axis,  y  may  be  calculated  by  multiplying  x 
by  the  tangent  of  the  half -angle. 


CHAP.  X] 


PARALLEL  FORCES 


169 


Problems 

1.  By  means  of  Equation  (9),  find  the  center  of  gravity  of  a  45-degree 
sector. 

2.  By  means  of  the  answer  of  Problem  1  find  the  center  of  gravity  of  a 
sector  of  which  the  angle  is  22°  30'. 

3.  Find  the  center  of  gravity  of  the  area  bounded  by  the  X  axis,  the 
curve  y  =  x^,  and  the  line  x  =  8  by  means  of  the  equations  of  this  article. 

4.  Find  the  center  of  gravity  of  a  segment  cut  off  by  a  line  at  a  distance 
from  the  center  equal  to  one-half  the  radius.     Calculate  the  area  and 
moment  of  the  half  sector  and  subtract  the  moment  and  area  of  the  triangle. 

5.  Find  the  center  of  gravity  of  the  area  bounded  by  the  X  axis,  the 
curve  y*  =  4z,  and  the  lines  x  =  4  and  x  =  9.     Find  the  moment  and  area 
of  the  entire  area  to  the  line  x  =  9  and  then  subtract  for  the  area  to  the 
Jeft  of  x  =  4. 

103.  Liquid  Pressure. — The  pressure  of  a  liquid  is  the  same  in 
all  directions,  and  is  proportional  to  the  depth  below  the  surface. 
If  w  is  the  weight  of  unit  volume  of  the  liquid,  and  y  the  depth  of  the 
surface  below  the  surface  of  the  liquid,  the  pressure  per  unit  area 
is  wy.  The  density  of  water  is 
nearly  62.5  pounds  per  cubic 
foot;  the  constant  w  for  water 
is  62.5  when  foot  and  pound 
units  are  used.  With  gram 
and  centimeter  units,  w  =  1. 

Figure  157  represents  a  tank 
of  liquid  in  which  there  is  a 
submerged  surface  MN.  An 
element  of  the  surface,  dA,  is 
located  at  a  vertical  distance 
y  below  the  surface  of  t;he 
liquid.  The  total  pressure  on 
one  side  of  the  surface  on  the  element  dA  is  wydA.  The  total 
pressure  on  the  entire  surface  is 

total  pressure  =  fwy  dA  =  wfydA.  &  (1) 

Since  J°ydA  is  the  moment  of  the  area  dA  with  respect  to  the 
surface  of  the  liquid 

fydA  =  yA,  (2) 

total  pressure  =  wyA.  (3) 

The  total  pressure  of  a  liquid  on  a  surface  is  the  weight  of  a 
column  of  liquid  whose  area  is  the  area  of  the  surface  and  whose 
height  is  the  depth  of  the  center  of  gravity  of  the  surface  below 
the  surface  of  the  liquid. 


FIG.  157. 


170 


MECHANICS 


[ART.    104 


Problems 

1.  One  end  of  the  tank  of  Fig.  157  is  6  feet  wide  and  8  feet  high  to  the 
surface  of  the  water.     Find  the  total  horizontal  pressure  on  this  vertical 
surface.  Arts.     12,000  Ib. 

2.  A  vertical  gate,  5  feet  wide  and  4  feet  high,  is  subjected  to  pressure  of 
water  which  rises  6  feet  above  the  top  of  the  gate.     Find  the  total  horizontal 
pressure  on  the  gate.  Ans.     10,000  Ib. 

3.  A  cylindrical  tank  with  axis  horizontal  is  6  feet  in  diameter.     It  is  filled 
with  water  and  connected  to  a  4-inch  pipe  in  which  water  stands  8  feet 
above  the  axis  of  the  cylinder.     Find  the  total  pressure  on  the  end. 

Ans.     14,137  Ib. 

4.  Solve  Problem  3  if  the  cylinder  is  half  full  of  water.     Ans.     1125  Ib. 
6.  A  tank  with  triangular  ends  is  6  feet  wide  at  the  top  and  4  feet  high. 

Find  the  pressure  at  one  end  when  the  tank  is  filled  with  water.     If  the 
tank  is  6  feet  long,  find  the  pressure  on  one  side. 

104.  Center  of  Pressure. — In  the  case  of  a  vertical  or  inclined 
surface,  the  pressure  varies  with  the  depth.     The  average  pressure 


is  that  at  the  center  of  gravity  of  the  area.  The  resultant  pres- 
sure is  below  the  center  of  gravity.  Fig.  158  shows  the  variation 
of  pressure  against  a  vertical  surface.  The  same  relation  holds 
for  an  inclined  surface.  Since  the  pressure  varies  as  the  depth 
below  the  surface,  it  may  be  represented  by  the  weight  of  a 
volume  of  liquid  between  two  planes  which  intersect  at  the  sur- 
face of  the  liquid.  In  the  case  of  a  rectangular  plane  surface, 
Fig.  158,  II,  the  pressure  is  equivalent  to  the  weight  of  a  wedge 
of  constant  width  equal  to  the  width  of  the  surface.  The  pressure 

at  the  bottom  is  wy.     The  average  pressure  is  -77,  which  is  the 

« 

pressure  at  the  center  of  gravity  of  the  area.  The  resultant  pres- 
sure is  located  at  the  center  of  gravity  of  the  wedge.  The  center 
of  gravity  of  the  wedge  is  the  same  as  that  of  a  triangular  plate, 
which  is  two-thirds  the  height  from  the  vertex.  The  position 


CHAP.  X]  PARALLEL  FORCES  171 

of  the  resultant  pressure  is  called  the  center  of  pressure.  The 
center  of  pressure  on  a  rectangular  surface  which  reaches  the  top 
of  the  liquid  is  two-thirds  the  height  of  the  surface  below  the 
surface  of  the  liquid.  This  distance  is  measured  parallel  to  the 
rectangular  surface. 

Problems 

1.  A  vertical  rectangular  gate  is  5  feet  wide  and  6  feet  high.     It  is  sub- 
jected to  pressure  of  water  which  just  reaches  the  top.     Find  the  total 
pressure  and  the  center  of  pressure.     If  the  gate  is  held  by  bolts  at  the  top 
and  bottom,  find  the  tension  in  each. 

Ans.     Tension  in  top  bolts  =  1875  lb.;  tension  in  bottom  bolts  =  3750  Ib. 

2.  A  vertical  rectangular  gate  is  5  feet  wide  and  12  feet  high.     It  is  sub- 
jected to  the  pressure  of  12  feet  of  water  on  one  side  and  the  pressure  of  6 
feet  of  water  on  the  other.     Find  the  force  at  the  top  and  bottom  required 
to  hold  the  gate  against  this  pressure.  Ans.     6562  lb.  at  the  top. 

3.  A  vertical  triangular  gate  is  8  feet  wide 
at^  the  bottom  and  12  feet  high.     It  is  sub- 
jected to  water  pressure  on  one  side.     Find 
the  center  of  pressure,  Fig.  158,  III. 

4.  A  vertical  rectangular  gate,  8  feet  wide 
and  12  feet  high,  is  subjected  to  the  pressure 
of  water  which  rises  4  feet  above  the  top. 
Find  the  horizontal  force  at  the  top  required 

to  hold  the  gate.     Where  is  the  center  of     ^OOl^/tt^        75O 
pressure?    Fig.  159. 

T  6.  Find  the  center  of  pressure  of  the  semicircular  area  of  Problem  4  of  the 
preceding  article. 

Generally,  the  location  of  the  center  of  gravity  of  the  wedge- 
shaped  solid  which  represents  the  pressure  is  not  known.  In 
Art.  153,  a  general  method  of  solution  is  given,  which  involves 
the  radius  of  gyration  of  the  area  of  the  surface.  The  determina- 
tion of  the  center  of  pressure  by  means  of  the  wedge  of  pressure 
is  convenient  for  rectangular  surfaces. 

105.  Summary.— The  resultant  of  parallel  forces  in  space  is 
equal  to  the  algebraic  sum  of  the  forces.  The  moment  of  the 
resultant  about  any  axis  is  equal  to  the  sum  of  the  moments  of 
the  separate  forces  about  that  axis. 

For  a  problem  of  equilibrium  of  parallel  forces  in  space  there 
are  three  unknowns  and  three  independent  equations.  These 
equations  may  be  two  moments  and  one  resolution  or  three 
moments. 

The  center  of  mass  is  the  point  of  application  of  the  resultant 
of  parallel  forces  which  are  proportional  to  the  masses  of  the 


172  MECHANICS  [ART.   105 

particles  upon  which  they  act.     Center  of  mass  is  practically 
the  same  as  center  of  gravity. 

Center  of  gravity  is  calculated  by  dividing  the  sum  of  the 
moments  by  the  sum  of  the  masses.  This  calculation  locates 
one  plane  through  the  center  of  gravity.  Two  other  planes  are 
found  in  the  same  way.  The  center  of  gravity  lies  at  the  inter- 
section of  the  three  planes. 

_  =  /Vdm 
fdm 
fy'dm 


_  =  fz'dm 
fdm  ' 

in  which  x,  y  and  z  are  coordinates  of  the  center  of  gravity  of 
the  body  and  x',  y',  and  z'  are  the  coordinates  of  the  elements  or 
particles  which  compose  the  body. 

When  the  density  is  constant,  the  elements  of  volume  may 
replace  the  elements  of  mass.  When  the  body  is  a  plate  of  uni- 
form thickness  and  density,  the  elements  of  area  may  be  used 
in  place  of  elements  of  mass. 

If  a  body  of  uniform  density  has  a  plane  of  symmetry,  the 
center  of  gravity  lies  in  that  plane. 

The  center  of  gravity  of  a  triangular  plate  of  uniform  density 
and  thickness  is  located  at  the  intersection  of  the  medians, 
which  is  at  a  distance  from  the  base  equal  to  one-third  of  the 
altitude  of  the  triangle. 

The  center  of  a  pyramid  or  cone  of  uniform  density  is  located 
at  a,  distance  from  the  base  equal  to  one-fourth  the  altitude. 

The  pressure  of  a  liquid  on  a  plane  surface  consists  of  parallel 
forces  perpendicular  to  the  surface.  This  pressure  is  directly 
proportional  to  the  vertical  depth  below  the  surface  of  the  liquid. 
The  average  pressure  on  any  surface  is  the  pressure  at  the  center 
of  gravity  of  the  surface.  The  point  of  application  of  the  resul- 
tant pressure  is  called  the  center  of  pressure.  The  center  of 
pressure  may  be  found  by  means  of  the  center  of  gravity  of  a 
wedge-shaped  solid.  The  center  of  pressure  on  a  rectangular 
plane  surface  which  extends  to  the  surface  of  the  liquid  is  at 
two-thirds  the  altitude  of  the  surface  below  the  surface  of  the 
liquid. 


CHAPTER  XI 
FORCES  IN  ANY  POSITION  AND  DIRECTION 

106.  Couples  in  Parallel  Planes. — In  Art.  70,  it  was  shown 
that  a  couple  may  be  balanced  by  another  couple  of  equal  magni- 
tude and  opposite  sign  acting  in  the  same  plane,  and  that  the 
forces  of  this  second  couple  might  have  any  magnitude,  direction, 
and  position  in  the  plane.  It 
will  now  be  shown  that  it  is 
not  necessary  for  the  couples 
to  be  in  the  same  plane.  A 
couple  may  be  balanced  by  a 
couple  in  any  parallel  plane, 
provided  the  moments  are 
equal  and  opposite.  Figure 
160  shows  a  wheel  to  which 
two  ropes  are  attached.  The 
ropes  run  horizontally  in  oppo- 
site directions.  If  a  force  P  is  FlG  160 
applied  to  the  lower  rope,  and 

if  the  upper  rope  is  attached  to  a  fixed  point,  the  tensions  in  the 
two  ropes  will  be  equal  and  opposite,  and  form  a  counter-clock- 
wise couple.  The  center  of  gravity  of  the  system  is  supposed  to 
be  located  at  W.  The  plane  of  the  weight  and  of  the  upward 
reaction  R  is  parallel  to  the  plane  of  the  wheel.  It  is  possible  to 
make  the  force  P  of  such  magnitude  that  the  system  is  in  equilib- 
rium. The  weight  W  and  the  reaction  R  form  a  clockwise 
couple,  which  is  equal  and  opposite  to  that  of  the  forces  in  the 
ropes. 

Figure  161  shows  two  forces,  PI  and  ?2,  each  of  magnitude  P. 
These  forces  are  at  a  distance  a  apart  in  the  same  vertical  plane. 
Suppose  that  an  opposite  couple  made  up  of  two  forces  Qi  and 
Q2  at  a  distance  b  apart  acts  in  a  parallel  plane.  If  the  moments 
are  equal  so  that 

Pa  =  Qb 

it  may  be  shown  that  these  couples  balance  and  produce 
equilibrium. 

173 


174 


MECHANICS 


[ART.   106 


FlG   161 


According  to  Art.  70,  the  couple  Qb  may  be  replaced  by  an 
equal  couple  Pa  in  its  plane.  In  Fig.  161,  these  forces  are 
P3  and  P4  and  are  parallel  to  PI  and  P2.  In  Fig.  161,  ABCD 
is  a  quadrilateral  perpendicular  to  the  forces.  Since  AB  =  a  = 
CD,  and  since  AB  is  parallel  to  CD,  the  quadrilateral  is  a  paral- 
lelogram. The  upward  force  PI  of  the  first  couple  and  the  up- 
ward force  P3  of  the  second  couple  may  be  replaced  by  a  force 
of  magnitude  2P  at  a  point  midway  between  them.  The  down- 
ward force  P2  and  the  downward 
force  ?4  may  be  replaced  by  their 
resultant  of  magnitude  2P  at  a  point 
midway  between  C  and  B.  The 
line  of  the  upward  resultant  passes 
through  the  middle  of  one  diagonal 
of  the  parallelogram,  and  the  line  of 
the  downward  force  passes  through 

the  middle  of  the  other  diagonal.  These  two  forces  must  lie  in 
the  same  line.  They  are  equal  and  opposite;  consequently,  they 
balance  and  produce  equilibrium. 

Since  two  couples  which  balance  the  same  couple  are  equiva- 
lent, two  couples  in  parallel  planes  which  are  numerically  equal 
and  have  the  same  sign  are  equivalent.  The  combined  moment  of 
two  couples  in  the  same  plane  is  the  algebraic  sum  of  the  moments 
of  the  separate  couples.  The  combined  moment  of  couples  in 
parallel  planes  is  the  algebraic  sum  of  the  moments  of  the  several 
couples. 

The  addition  of  couples  in  parallel  planes  may  be  proved 
in  a  slightly  different  way.  In  Fig.  161,  suppose  that  both 
couples  are  clockwise  so  that  the 
force  P4  is  up  and  the  force  P3  is 
down.  The  resultant  of  PI  and  P4 
is  a  force  2P  midway  between  them. 
The  resultant  of  P2  and  P3  is  an 
equal  downward  force.  These  two 
resultants  are  at  a  distance  a  apart, 
and  form  a  couple  of  magnitude 
2Pa.  This  couple  is  in  a  plane 
parallel  to  the  planes  of  the  original  couples  and  midway 
between  them. 

Figure  162  applies  to  unequal  couples.  It  will  be  shown  that 
the  resultant  couple  may  be  in  any  plane  parallel  to  the  plane  of 


P 


./[<___ 


,         -->| 


1-9 


CHAP.  XI]  FORCES  IN  POSITION  AND  DIRECTION   175 

the  couples  to  be  added.  This  figure  represents  a  plane  perpen- 
dicular to  the  planes  of  the  two  couples.  One  of  these  couples 
is  made  up  of  a  force  P  upward  perpendicular  to  the  plane  of 
the  paper  and  an  equal  force  downward  at  a  distance  a  from 
this  upward  force.  In  the  second  couple  the  forces  are  Q  down- 
ward at  the  right  and  an  equal  force  upward  at  a  distance  6. 
The  distance  between  the  planes  of  the  two  couples  is  d.  The 
resultant  of  the  upward  parallel  forces  P  and  Q  is  a  force  P  +  Q 
at  a  distance  x  from  the  plane  of  the  first  couple.  By  moments 

•  - 


The  resultant  of  the  two  downward  forces  is  a  downward 
force  P  +  Q  at  the  same  distance  from  the  plane  of  the  first 
couple.  The  moment  arm  of  the  resultant  couple  is  c. 


Resultant    moment  =  (P  +  Q)c  =  Pa  +  Qb.  (4) 

The  couple  Qb  may  be  replaced  by  any  other  couple  in  its 
plane,  provided  the  moment  is  the  same  in  magnitude  and  sign. 
If  Q  were  made  larger  and  b  smaller,  the  plane  of  the  resultant 
couple  would  be  closer  to  the  couple  Qb.  By  changing  the  rela- 
tive values  of  Q  and  6,  it  may  be  shown  that  the  resultant  couple 
may  come  in  any  plane  parallel  to  the  planes  of  the  original 
couples. 

An  ordinary  windlass  affords  an  example  of  the  equilibrium 
of  parallel  couples.  The  force  on  the  crank  and  part  of  the 
reactions  of  the  bearings  form  one  couple.  The  load  on  the  rope 
and  the  remainder  of  the  reactions  of  the  bearings  form  the  other 
couple.  This  arrangement,  and  the  similar  one  of  pulleys  on 
shafting  are  complicated  by  the  fact  that  the  reactions  at  the 
bearings  are  not  usually  in  the  same  plane  as  the  turning  forces, 
and  must  be  resolved  into  components  to  get  a  pair  of  couples. 

107.  A  Couple  as  a  Vector.  —  It  has  been  shown  that  a  couple 
may  be  replaced  by  another  couple  of  the  same  magnitude  and 
sign.  The  forces  of  the  couple  may  have  any  magnitude  and 
position,  and  may  be  located  anywhere  in  the  plane  of  the  original 
couple  or  in  any  plane  parallel  to  the  plane  of  the  original  couple. 


176  MECHANICS  [ART.  108 

The  easiest  way  to  express  the  direction  of  a  plane  is  by  means  of 
the  direction  of  a  line  normal  to  it.  All  parallel  planes  have  the 
same  normal.  A  couple  may  be  represented  by  a  line  normal  to 
its  plane.  The  length  of  the  line  gives  the  magnitude  of  the 
couple.  The  direction  of  the  line  bears  the  same  relation  to  the 
direction  of  rotation  of  the  couple  as  the  direction  of  motion  of  a 
right-handed  screw  bears  to  its  direction  of  rotation.  A  clock- 
wise couple  is  represented  by  a  vector  away  from  the  observer, 
a  counter-clockwise  couple  by  a  vector  toward  the  observer. 

A  vector  partly  represents  a  force,  since  it  gives  its  direction  and 
magnitude,  but  does  not  give  its  location.  A  vector  completely 
represents  a  couple,  since  the  forces  of  the  couple  may  be  any- 
where in  a  given  plane  or  in  parallel  planes. 

That  a  force  may  be  represented  by  a  vector,  and  that  the 
resultant  of  two  forces  is  given  by  their  vector  sum  has  been 
accepted  as  an  axiom  which  has  been  amply  verified  by  experi- 
ment. If  two  couples  which  are  not  in  parallel  planes  be  repre- 
sented by  vectors,  and  if  it  may  be  shown  that  the  resultant  of 
these  two  couples  is  a  third  couple  which  is  represented  fully  by 
the  vector  sum  of  these  vectors,  then  the  assumption  that  a 
couple  is  a  vector  may  be  regarded  as  valid.  This  proof  will  be 
given  in  Art.  108. 

108.  Resultant  Couple. — Figure  163,  I,  represents  two  planes 
which  intersect  on  the  line  A  B.  A  counter-clockwise  couple  of 
moment  Pa  is  supposed  to  be  acting  in  the  plane  CAB  and  a  second 
counter-clockwise  couple  Qb  in  the  plane  DAB.  The  couple  Pa 
may  be  anywhere  in  its  plane.  One  force  P2  may  be  regarded 
as  acting  along  the  line  A  B  and  the  second  force  PI  at  a  dis- 
tance a  from  that  line.  The  couple  Qb  may  be  replaced  by  a 
couple  PC  in  its  plane,  provided  PC  =  Qb.  One  force  of  this  new 
couple  may  be  regarded  as  acting  in  the  line  AB  in  a  direction 
opposite  to  the  force  ?2.  The  remaining  force  will  then  be  in  the 
plane  DAB  at  a  distance  c  from  the  line  AB.  The  forces  ?2 
and  ?4  acting  along  the  line  AB  in  opposite  directions  balance 
each  other.  The  remaining  forces  PI  and  ?4  form  a  new  counter- 
clockwise couple.  If  d  is  the  distance  from  PI  to  ?4,  the  moment 
of  this  couple  is  Pd.  Figure  163,  II,  is  a  space  triangle  in  a  plane 
perpendicular  to  the  planes  CAB  and  DAB.  The  line  HL  of 
length  d  in  this  triangle  is  the  distance  between  the  forces  of  the 
resultant  couple.  Figure  163,  III,  is  a  vector  diagram.  The  line 
EF  of  length  Pa  represents  the  original  couple  in  the  plane  CAB, 


CHAP.  XI]   FORCES  IN  POSITION  AND  DIRECTION    177 

The  line  EF  is  normal  to  this  plane,  and  therefore,  lies  in  the  plane 
of  the  triangle  HLK,  and  is  perpendicular  to  the  line  HK.  The 
line  FG  is  normal  to  the  plane  DAB  and  is  normal  to  the  line  KL 
of  the  space  triangle.  Since 

Pa  _  a 
PC  ~  ~c' 

and  since  the  angle  at  F  is  equal  to  the  angle  at  K,  the  space 
diagram  and  the  vector  diagram  are  similar  triangles.  As  the 


FIG.  163. 


angle  at  E  is  equal  to  the  angle  at  H,  EG  is  normal  to  HL.     Since 
the  sides  are  proportional 

EG  =  d 

Pa    ''     a' 


If  two  couples  are  represented  by  vectors  normal  to  their 
respective  planes,  the  resultant  of  these  couples  is  represented 
by  the  vector  sum  of  the  two  vectors. 

Example 

A  force  of  12  pounds  is  vertically  downward.  Another  force  of  12  pounds 
vertically  upward  is  located  2  feet  east  of  this  first  force.  A  second  couple 
is  made  up  of  a  horizontal  force  of  8  pounds  directed  north  and  an  equal 
force  directed  south  at  a  distance  of  4  feet  above  this  force.  Find  the 
resultant  couple, 
12 


178 


MECHANICS 


[ART.   109 


The  first  force  has  a  moment  of  24  foot-pounds.  It  is  represented  by  a 
horizontal  vector  24  units  in  length  directed  south.  The  second  couple 
is  represented  by  a  horizontal  vector  32  units  in  length  directed  east.  The 
resultant  couple  is  represented  by  their  vector  sum.  This  is  a  horizontal 
vector  40  units  in  length  directed  south  53°  08'  east.  The  resultant  couple 
is  in  a  vertical  plane  normal  to  this  resultant  vector. 


Problem 

1.  A  box  is  6  feet  long  from  east  to  west,  4  feet  wide,  and  3  feet  high.  It 
weighs  160  pounds  and  its  center  of  mass  is  at  its  center.  At  the  northwest 
corner  at  the  bottom,  the  box  rests  upon  a  support  which  permits  it  to  turn 
but  not  to  move  laterally.  A  horizontal  force  of  60  pounds  directed  south 
is  applied  at  the  south  east  corner  at  the  bottom.  A  horizontal  force  of  80 
pounds  directed  east  is  applied  at  the  northwest  corner  at  the  top.  In 
what  plane  will  these  three  forces  and  the  reactions  at  the  support  turn  the 
box,  and  what  is  the  resultant  moment? 

When  the  planes  of  the  couples  which  are  to  be  combined  are 
not  at  right  angles  to  each  other,  it  is  advisable  to  resolve  each 
vector  along  three  coordinate  axes;  then  combine  the  components 
in  exactly  the  same  way  as  was  used  in  the  calculation  of  the 
resultant  of  non-ooplanar  forces. 

Problem 

2.  In  Fig.  164,  find  the  sum  of  the  com- 
ponents of  the  moment  vectors  along  each  of 
the  three  axes. 

Ans.     170    ft.-lb.    about   a   vertical   axis; 
120    ft.-lb.  about    a   horizontal  axis  toward 
the  left;   150  ft.-lb.  about  a  horizontal  axis 
FIG.  164.  toward  the  front. 

109.  Forces  Reduced  to  a  Force  and  a  Couple. — In  Chapter 
VI  it  was  shown  that  a  set  of  non-concurrent,  coplanar  forces 
may  be  reduced  to  a  single  force  and  a  single  couple  in  the  same 
plane.  It  will  now  be  shown  that  a  set  of  non-concurrent  forces, 
which  are  not  coplanar,  may  likewise  be  reduced  to  a  single  force 
and  a  single  couple.  The  force  and  the  couple  will  not,  in  general, 
be  in  the  same  plane. 

In  Fig.  165,  the  line  AB  represents  the  line  of  a  single  force  of 
magnitude  P.  This  force  may  be  replaced  by  an  equivalent 
force  P  at  some  point  0  and  a  couple  in  the  plane  A  OB.  The 
moment  of  this  couple  is  the  product  of  the  force  P  multiplied 
by  the  distance  from  0  to  its  line  of  action.  A  second  force  Q 


CHAP.  XI]  FORCES  IN  POSITION  AND  DIRECTION    179 


(not  shown  in  the  drawing  may  likewise  be  replaced  by  a  force 
Q  at  0  and  a  couple  in  the  plane  through  0  and  its  line  of  action. 
The  resolution  of  any  number  of  forces  in  this  way,  gives  a  set 
of  concurrent  forces  at  0  and  a  set  of  couples.  The  concurrent, 
non-coplanar  forces  at  0  may  be  combined  into  a  single  resultant 


Y 


FIG.  165. 

by  the  methods  of  Art.  89.     The  couples  may  be  combined  into 
a  single  couple  by  a  similar  addition  of  vectors. 

In  the  case  of  non-concurrent,  coplanar  forces,  the  single  force 
and  the  single  couple  may  be  further  reduced  to  a  single  force, 
except  when  the  resultant  force  is  zero.  When  the  resultant 
force  is  zero,  the  final  result  is  a  couple.  In  the  case  of  non-con- 
current, non-coplanar  forces,  the  force  and  the  couple  can  not, 
in  general,  be  combined  into  a  single  force.  In  Fig.  166,  P  repre- 
sents a  force  through  the  point  0, 
and  Qb  represents  a  couple  in 
a  different  plane.  One  of  the 
forces  of  the  couple  may  be  made 
to  intersect  the  force  P  at  0. 
By  changing  the  value  of  6,  the 
force  Qi  may  be  given  any  de- 
sired  magnitude.  If,  however,  the 
couple  does  not  lie  in  a  plane  parallel  to  the  direction  of  the 
force  P,  the  two  intersecting  forces  can  not  be  made  to  balance. 
In  general,  the  resultant  of  the  forces  P  and  Qi  will  be  a  force  R. 
This  force  R  is  not  in  the  same  plane  with  the  force  Q2.  The  set 
of  forces  reduces  to  a  force  and  a  couple,  and  the  force  and  the 


R-«-^ 

9,      | 

0            1 

c 

\ 
FIG.  166. 

180  MECHANICS  [ART.  110 

couple  reduce  to  two  forces,  which  are  not  in  the  same  plane  and 
do  not  intersect. 

By  changing  the  distance  b,  the  value  of  Q  is  changed,  hence 
the  magnitude  and  position  of  the  force  Q2  and  the  magnitude  and 
direction  of  the  force  R  may  be  changed. 

110.  Equilibrium  of  Non-concurrent,  Non-coplanar  Forces. — 
It  has  been  shown  that  a  set  of  non-concurrent,  non-coplanar 
forces  may  be  reduced  to  a  single  force  through  some  arbitrarily 
chosen  point  and  a  single  couple.  The  magnitude  and  direction 
of  the  force  are  determined  by  the  conditions  of  the  problem. 
The  magnitude  and  direction  of  the  couple  depend  upon  the 
conditions  of  the  problem  and  the  location  of  the  arbitrarily 
chosen  point  through  which  the  single  force  acts.  To  specify 
fully  the  single  force  through  a  known  point,  three  factors  are 
required.  These  may  be  the  magnitude  of  the  force  and  two 
angles,  or  the  components  along  each  of  three  axes.  To  specify 
fully  the  couple  as  a  vector,  three  factors  are  required.  These 
may  be  one  magnitude  and  two  directions,  or  three  components. 
A  problem  of  non-concurrent,  non-coplanar  forces  in  equilibrium 
may,  therefore,  involve  six  unknowns  and  require  six  equations 
for  its  solution. 

For  equilibrium,  the  resultant  force  must  be  zero.  The  force 
diagram  for  all  the  forces  at  the  arbitrarily  chosen  point  forms  a 
closed  polygon  in  space.  Each  force  at  this  arbitrary  point  is 
identical  in  magnitude  and  direction  with  the  force  which  it 
replaces.  If  all  the  non-concurrent  forces  of  the  problem  are 
regarded  as  concurrent  and  combined  into  a  single  force  polygon, 
this  polygon  is  closed.  The  sum  of  the  components  along  each 
of  three  axes  is  zero. 

The  resultant  couple  must  also  be  equal  to  zero.  The  forces 
and  reactions  on  the  rigid  body  combine  to  make  up  this  resul- 
tant couple.  In  order  that  the  couple  may  be  zero,  it  is  necessary 
that  the  sum  of  the  moments  about  any  axis  be  equal  to  zero. 

To  solve  a  problem  of  equilibrium  in  which  there  are  six 
unknowns,  it  is  necessary  to  write  six  independent  equations. 
Three  of  these  must  be  moment  equations.  The  remaining 
three  may  be  either  resolution  equations  or  moment  equations. 
It  is  advisable  to  begin  with  a  moment  equation  about  some  axis 
which  will  eliminate  most  of  the  unknowns. 

In  the  special  instance  in  which  all  the  forces  are  parallel, 
the  number  of  unknowns  is  reduced  to  three.  Since  the  forces 


CHAP.  XI]  FORCES  IN  POSITION  AND  DIRECTION    181 

are  all  in  the  same  direction,  only  one  resolution  equation  may 
be  written.  The  other  two  equations  must  be  moment  equations. 
If  all  the  forces  are  in  parallel  planes,  the  resolution  perpen- 
dicular to  these  planes  will  give  zero  for  each  force.  In  this  case 
there  may  be  only  two  resolutions.  The  total  number  of  un- 
knowns is  reduced  to  five. 

Example 

A  horizontal  trap  door,  Fig.  167,  is  8  feet  long  and  6  feet  wide.  It  is 
supported  by  two  hinges,  each  1  foot  from  a  corner,  on  an  8-foot  edge. 
It  is  lifted  by  a  rope  attached  to  the  other  8-foot  edge  at  a  distance  of  2  feet 


Space  Diagram 


^H'-- 
•re '      11 

Force  Diagram 
FIG.  167. 


from  one  corner.  This  rope  passes  over  a  smooth  pulley  which  is  4  feet  above 
the  center  of  the  door.  The  door  weighs  60  pounds  and  its  center  of  mass 
is  at  the  center.  Find  the  tension  in  the  rope  and  the  horizontal  and  vertical 
components  of  the  hinge  reaction,  assuming  that  the  hinges  are  so  con- 
structed that  the  horizontal  force  parallel  to  their  line  is  all  taken  by  the 
left  hinge. 

The  unknown  quantities  are  the  tension  in  the  rope,  the  horizontal  and 
vertical  components  of  the  reaction  at  the  right  hinge,  two  horizontal 
components  and  one  vertical  component  at  the  left  hinge.  The  tension 
of  the  rope  is  made  up  of  three  components.  Since  the  direction  of  the 
rope  is  known,  the  problem  is  fully  solved  when  one  of  these  components 
is  determined. 

A  moment  equation  about  the  line  of  the  hinges  eliminates  five  of  the 
unknowns,  together  with  the  horizontal  components  of  the  tension  of  the 


182  MECHANICS  [ART.  110 

rope.     The  forces  which  have  moment  about  this  line  are  the  weight  of 
the  door  and  the  vertical  component  of  the  tension  in  the  rope. 

60  X  3  =  Vi  X  6, 
Vi  =  30  Ib. 

The  tension  in  the  rope  and  its  horizontal  components  may  now  be  com- 
puted by  resolutions.  (These  resolutions  form  no  part  of  the  six  equations, 
since  the  door  is  not  under  consideration  as  the  free  body.)  The  space 
diagram,  Fig.  167,  II,  is  similar  to  the  force  diagram,  Fig.  167,  III.  All 
the  dimensions  of  the  space  diagram  are  known.  The  vertical  force  Vi 
of  the  force  diagram  is  also  known.  The  other  forces  may  be  found  from 
the  geometry  of  the  similar  solids.  If  P  is  the  tension  in  the  rope, 

P_  =  /29. 
30         4  ' 
P  =  40.39  Ib. 

Similarly,  HI  =  15  Ib.,  H2  =  22.5  Ib.     These  components  are  convenient 
in  finding  the  components  at  the  hinges. 

Taking  moments  about  an  axis  which  passes  through  the  left  hinge  and  is 
parallel  to  the  6-foot  edges, 

F3  X  6  =  60  X  3  -  30  X  5; 
F3  =  5  Ib. 

Taking  moments  about  an  axis  through  the  right  hinge  parallel  to  the 
6-foot  edges, 

F2  X  6  =  60  X  3  -  30  X  1; 
F2  =  25  Ib. 

Check  by  a  vertical  resolution, 

Fi  +  F2  +  F3  =  30  +  25  +  5  =  60. 

To  find  H4,  take  moments  about  a  vertical  axis  through  the  right  hinge. 
This  eliminates  H3  and  H8. 

H4  X  6  =  tf !  X  6  +  H*  X  1 

15      X  6  =  90 
22.5  X  1  =  22.5 
6#4  =  112.5, 
Hi  =  18.751b. 

To  find  H3,  take  moments  about  a  vertical  axis  through  the  left  hinge, 

15      X  6  =    90  counter-clockwise, 
22.5  X  5  =  112. 5  clockwise, 

6#3  =  22.5  clockwise, 
ffa  =  3.75lb. 

The  moment  of  H3  about  the  vertical  axis  through  the  left  hinge  must 
balance  a  clockwise  moment.     The  direction  of  H3  must  be  opposite  the 


CHAP.  XI]  FORCES  IN  POSITION  AND  DIRECTION    183 

direction  of  the  arrow  in  Fig.  167.     H3  might  have  been  computed  by  a 
horizontal  resolution.     Resolving  as  a  check, 

3.75  +  18.75  =  22.5. 

By  resolutions  parallel  to  the  line  of  the  hinges, 
#5  =  Hl  =  15  Ib. 

The  problem  has  been  solved  by  five  moment  equations  and  one  resolution 
equation,  and  partly  checked  by  two  resolution  equations.  Three  of  the 
moments  were  taken  about  horizontal  axes  and  two  were  taken  about  vertical 
axes. 

Problems 

1.  A  horizontal  trap  door,  similar  to  Fig.  167,  is  10  feet  long,  7  feet  wide, 
weighs  80  pounds,  and  has  its  center  of  mass  at  the  center.  It  is  hinged 
1  foot  from  the  corners  on  one  long  side  and  lifted  by  a  rope  at  one  corner  of 
the  opposite  side.  The  rope  passes  over  a  smooth  pulley  which  is  8  feet 
above  the  center  of  the  door.  Find  all  the  reactions. 


Diagram 


Vector  Diagram  of  Coup/es 
FIG.  168. 


2.  A  horizontal  plank,  24  inches  long,  18  inches  wide,  weighs  18  pounds 
and  has  its  center  of  gravity  at  the  center.  It  is  supported  at  the  middle 
of  one  24-inch  side,  and  2  inches  from  one  corner  and  4  inches  from  the  other 
corner  of  the  other  24-inch  side.  Find  the  reactions  of  these  supports. 

Ans.     9  Ib.,  4  Ib.,  5  Ib. 

.  3.  In  problem  2  consider  the  downward  force  of  18  pounds  as  made  up 
of  9  pounds,  4  pounds,  and  5  pounds,  and  consider  that  these  forces  form 
couples  with  the  three  reactions.  The  vectors  which  represent  these  couples 
lie  in  the  plane  of  the  board.  Draw  the  vector  diagram  for  the  couples, 
Fig.  168. 

4.  In  Fig.  168,  the  reaction  of  9  pounds  is  moved  4  inches  to  the  right  of 
the  middle.  Find  the  other  reactions  and  draw  the  vector  diagram  for  the 
three  couples  in  equilibrium. 

6.  The  mast  of  a  stiff-leg  derrick,  Fig.  169,  is  30  feet  high.  The  stiff  legs 
are  in  vertical  planes  at  right  angles  to  each  other.  One  of  these  planes 
extends  south  and  the  other  extends  west  of  the  mast.  The  stiff  legs  make 
angles  of  45  degrees  with  the  vertical.  The  boom  is  40  feet  long  and  carries 


184  MECHANICS  [ART.   110 

a  load  of  2,000  pounds  on  the  end.  Find  the  tension  in  the  stiff  legs  and 
the  reactions  at  the  bottom  of  the  mast  resulting  from  this  load  when  the 
boom  is  elevated  20  degrees  above  the  horizontal  in  a  vertical  plane  which 
is  north  50  degrees  east. 

Regard  the  boom,  mast,  and  the  cables  which  connect  them  as  a  free 
body  in  equilibrium.  Begin  by  taking  moments  about  a  horizontal  north 
and  south  line  through  the  bottom  of  the  mast.  This  eliminates  the  reac- 
tions at  the  bottom  of  the  mast  and  the  tension  in  the  stiff  leg  in  the  north 
and  south  vertical  plane.  Let  Vi  be  the  vertical  component  of  the  tension 
in  the  west  stiff  leg.  Consider  this  component  applied  at  the  base  of  the 
leg,  where  the  moment  arm  of  the  horizontal  component  is  zero. 

Vi  X  30  =  2000  X  40  cos  20°  cos  40°  =  57,585  ft.-lb. 

Vi  =  1919  Ib. 
The  tension  in  the  west  stiff  leg  =  Vi  sec  45°  =  2714  Ib. 


Find  the  tension  in  the  other  stiff  leg  by  moments  in  a  similar  manner 
and  find  the  three  components  of  the  reaction  at  the  base  by  three  resolu- 
tions. Check  by  moments. 

There  are  only  five  unknowns.  The  only  forces  which  are  not  in  the  plane 
of  the  mast  and  boom  are  the  forces  which  act  on  the  mast.  If  there  were 
transverse  forces  acting  on  the  boom,  there  could  be  another  reaction  with 
a  horizontal  component.  This  would  be  the  sixth  unknown. 

6.  Solve  Problem  5  when  the  boom  is  south  10  degrees  east.     What  is 
the  direction  of  the  vertical  reaction  at  the  bottom  of  the  mast? 

The  student  should  look  up  a  stiff-leg  derrick.  What  provision  is  made 
for  the  reversed  reaction  at  the  bottom  of  the  mast  when  the  boom  is  longer 
than  the  horizontal  projection  of  the  legs?  What  provision  is  made  for 
resisting  the  vertical  component  of  the  tension  in  the  legs? 

7.  A  derrick  mast  is  40  feet  long.     The  boom  is  50  feet  long,  weighs  1000 
pounds,  and  has  its  center  of  mass  at  the  center.     One  guy  rope  is  west  and 


CHAP.  XI]  FORCES  IN  POSITION  AND  DIRECTION   185 

another  is  north  10  degrees  east.  These  ropes  make  angles  of  30  degrees 
with  the  horizontal.  The  boom  is  horizontal  in  a  position  south  60  degrees 
east  and  carries  a  load  of  2000  pounds.  Find  the  tension  in  the  west  guy 
rope  by  moments  about  a  horizontal  line  through  the  bottom  of  the  mast 
and  the  bottom  of  the  other  guy  rope.  Ans.  3443  Ib. 

8.  At  what  position  of  the  boom  in  Problem  7  will  the  tension  in  the  west 
guy  rope  be  the  greatest,  and  what  will  be  its  value? 

Ans.  3664  Ib.  when  the  boom  is  south  80  degrees  east. 

9.  Find  the  four  remaining  unknowns  in  Problem  7. 

10.  A  windlass,  Fig.  170,  is  1  foot  in  diameter  and  4  feet  long  between 
bearings.     The  crank  is  2  feet  long  from  the  center  of  the  crank  pin  to  the 
axis  of  the  axle.     The  force  is  applied  to  the  crank  pin  at  a  distance  of  1  foot 
from  the  plane  through  the  right  bearing  perpendicular  to  the  axis.     A  rope 
wound  round  the  windlass  20  inches  from  the  left  bearing  carries  a  load  of 
480  pounds.     The  crank  makes  an  angle  of  30  degrees  to  the  right  of  the 


FIG.  170. 

vertical  upward  viewed  from  the  right.  The  force  on  the  crank  is  normal 
.to  the  plane  through  the  axis  of  the  windlass  and  the  crank  pin.  Find  this 
force  and  the  horizontal  and  vertical  reactions  at  the  bearings.  The  forces 
are  all  in  planes  perpendicular  to  the  axis  so  that  there  are  only  five  un- 
knowns. 

Ans.     P  =  120  Ib.;  Vi  =  265  Ib.;  72  =  275  Ib.;  Hi  =  25.98  Ib.;  H2  = 
129.90  Ib. 

11.  Solve  Problem  10  when  the  crank  is  45  degrees  to  the  right  of  the 
vertical. 

12.  A  shaft  is  6  feet  long  from  center  to  center  of  bearings  and  weighs 
200  pounds.     At  one  foot  from  the  left  bearing  it  carries  a  pulley  4  feet  in 
diameter  which  weighs  240  pounds.     At  2  feet  from  the  right  bearing  it 
carries  a  pulley  3  feet  in  diameter  which  weighs  180  pounds.     The  belt  from 
the  left  pulley  runs  forward  horizontally  and  exerts  a  pull  of  600  pounds 
at  the  top  and  150  pounds  at  the  bottom.     The  belt  from  the  right  pulley 
passes  around  a  pulley  6  feet  in  diameter  on  a  second  shaft.     The  axis  of 
this  second  shaft  is  10  feet  below  the  axis  of  the  first  shaft.     The  tension 
in  the  front  of  this  belt  is  200  pounds.     Find  the  tension  in  the  other  part  of 
the  belt  and  the  components  of  the  reactions  of  the  bearings. 


186  MECHANICS  [ART.   110 

13.  A  door  is  10  feet  high,  12  feet  wide,  weighs  180  pounds,  and  has  its 
center  of  gravity  at  the  center.     It  is  hinged  1  foot  from  the  top  and  1  foot 
from  the  bottom.     The  top  hinge  is  6  inches  north  of  a  vertical  line  through 
the  bottom  hinge.     The  hinges  are  so  placed  that  the  lower  hinge  takes  all 
the  force  parallel  to  the  edge  of  the  door.     Find  the  force  on  each  hinge 
when  the  door  is  directly  south  of  the  hinges.     The  door  is  turned  until  the 
top  and  bottom  are  east  and  west  and  is  held  by  a  horizontal  force  applied 
to  the  edge  4  feet  from  the  bottom.     Find  this  force  and  the  components 
of  the  forces  of  the  hinges. 

14.  A  rod  12  feet  long,  weighing  270  pounds,  with  its  center  of  mass  at 
the  center,  is  hung  horizontally  north  and  south  and  supported  by  three 
cords.     One  cord  is  attached  2  feet  from  the  north  end.     The  others  are 
attached  2  feet  from  the  south  end.     One  of  the  latter  makes  an  angle  of 
34  degrees  within  the  vertical  in  a  vertical  plane  which  is  north  75  degrees 
east.     The  other  makes  an  angle  of  45  degrees  with  the  vertical  in  a  vertical 
plane  which  is  south  60  degrees  west.     Find  the  direction  of  the  cord  near 
the  north  end  and  the  tension  in  each  of  the  three. 

Ans.     6°  34'  north  of  the  vertical,  tension  136.9  lb.;  92.9  lb.;  81.9  Ib. 

15.  A  bar,  weighing  114  pounds,  with  its  center  of  mass  at  the  middle, 
is  supported  in  a  north  and  south  horizontal  position  by  means  of  three  ropes. 
One  rope  is  attached  5  feet  north  of  the  middle.     The  second  rope  is  attached 
2  feet  south  of  the  middle.     The  upper  end  of  this  rope  is  fastened  to  a 
point  which  is  5  feet  above,  2  feet  south  and  3  feet  east  of  the  point  of  attach- 
ment to  the  bar.     The  third  rope  is  attached  7  feet  south  of  the  middle  of 
the  bar.     The  upper  end  of  this  rope  is  fastened  to  a  point  which  is  6  feet 
above,  4  feet  west  and  1  foot  north  of  the  point  of  attachment  to  the  bar. 
Find  the  tension  in  each  rope  and  the  direction  of  the  first  one. 

Ans.     First  rope  is  19°  02'  with  the  vertical  in  a  vertical  plane  which  is 
north  38°  40'  west.     The  tensions  are  51.44  lb.,  52.84  lb.,  and  27.30  lb. 

16.  A  wheelbarrovT  is  5  feet  long  from  axle  to  handles,  and  2  feet  wide 
between  handles.     A  *oad  of  180  pounds  is  placed  3  inches  to  the  right  of 
the  center  and  12  inches  from  the  axis  of  the  axle.     The  wheelbarrow  is 
held  horizontal  and  pushed  up  an  inclined  plane  which  makes  an  angle  of  5 
degrees  with  the  horizontal.     Find  the  reaction  at  the  wheel  and  the  direc- 
tion and  magnitude  of  the  force  at  each  handle. 

17.  A  cylinder  6  inches  in  diameter  and  weighing  60  pounds  is  suspended  in 
a  horizontal  position  by  means  of  two  ropes.     One  rope  supports    the 
cylinder  2  feet  from  the  middle.     This  rope  passes  under  the  cylinder,  and 
both  ends  leave  the  cylinder  in  the  samevertical  plane  perpendicular  to  its  axis. 
One  end  makes  an  angle  of  45  degrees  with  the  vertical  and  the  other  makes 
an  angle  of  30  degrees  with  the  vertical.     The  friction  is  sufficient  to  prevent 
the  cylinder  from  slipping.     The  second  rope  is  wound  round  the  cylinder 
and  fastened  to  it.     How  far  must  this  second  rope  be  placed  from  the  middle 
of  the  cylinder?     Solve  graphically,  neglecting  the  diameter  of  the  ropes. 

18.  A  right  cone,  2  feet  in  diameter  at  the  base  and  8  feet  high,  is  supported 
with  its  axis  horizontal  by  means  of  a  vertical  rope  which  is  wound  round  it 
several  times  but  is  not  fastened  to  it.     The  friction  is  sufficient  to  prevent 
slipping.     One  end  of  the  rope  leaves  the  cone  4  inches  from  the  base. 
Where  must  the  other  end  leave?    Solve  graphically. 


CHAP.  XI]   FORCES  IN  POSITION  AND  DIRECTION   187 

111.  Summary. — Couples  in  parallel  planes  which  have  the 
same  magnitude  and  sign  are  equivalent.  If  their  signs  are 
opposite,  they  balance  each  other.  Couples  in  parallel  planes 
are  added  algebraically. 

A  couple  may  be  represented  by  a  vector  perpendicular  to  its 
plane.  The  magnitude  of  the  vector  is  proportional  to  the 
moment  of  the  couple.  The  direction  of  the  vector  bears  the 
same  relation  to  the  direction  of  rotation  of  the  couple  as  the  direc- 
tion of  motion  of  a  right-handed  screw  bears  to  its  direction  of 
rotation.  (This  is  merely  a  convention.  The  opposite  direction 
might  have  been  chosen  just  as  well.) 

A  set  of  non-concurrent,  non-coplanar  forces  acting  on  a  rigid 
body  may  be  reduced  to  a  single  force  at  any  arbitrarily  chosen 
point  and  a  single  couple.  This  single  force  is  equivalent  to  the 
resultant  of  all  the  forces  which  act  on  the  body.  The  force  and 
the  couple  may  be  reduced  to  two  forces.  These  two  forces  do 
not,  in  general,  intersect. 

Any  rigid  body  may  be  held  in  equilibrium  by  a  single  force 
which  is  equal  and  opposite  to  the  resultant  of  the  applied  forces 
and  a  couple.  Any  rigid  body  may  be  held  in  equilibrium  by 
two  forces  of  proper  magnitude,  direction,  and  location. 

The  conditions  of  equilibrium  are: 

(1)  The  force  polygon  must  close. 

(2)  The  sum  of  the  moments  about  any  axis  must  be  zero. 
The  condition  that  the  force  polygon  must  close  is  expressed 
algebraically  by  the  statement  that  the  sum  of  the  components 
along  any  axis  is  equal  to  zero.     By  taking  resolutions  along  three 
axes,  not  more  than  two  of  which  are  in  the  same  plane,  three 
independent    equations    are    obtained.     By    taking    moments 
about  three  axes,  three  more  independent  equations  are  obtained. 

A  problem  of  the  equilibrium  of  non-concurrent,  non-coplanar 
forces  may  involve  six  unknowns  and  six  independent  equations 
of  mechanics.  There  may  be  other  equations  depending  upon 
the  geometric,  algebraic,  or  other  conditions  of  the  problem. 
These  six  equations  may  be  three  resolutions  and  three  moments. 
Any  or  all  of  the  resolutions  may  be  replaced  by  moment 
equations. 

Many  problems  involve  fewer  than  six  unknowns.  In  Problem 
2  of  Art.  110,  all  the  forces  are  parallel.  One  resolution  com- 
pletely specifies  these  forces,  hence  two  resolution  equations  must 
be  omitted.  The  forces  exert  no  moment  about  a  vertical  axis. 


188  MECHANICS  [ART.  ill 

This  eliminates  one  possible  moment  equation.  When  non- 
coplanar  forces  are  parallel,  the  number  of  unknowns  is  reduced 
to  three.  In  Problem  5  of  Art.  110,  there  is  no  moment  about 
a  vertical  axis.  This  eliminates  one  couple  and  reduces  the 
possible  unknowns  to  five.  In  problem  10  of  Art.  110,  the  forces 
have  no  components  parallel  to  the  axis  of  the  cylinder.  This 
eliminates  a  resolution  equation  and  reduces  the  unknowns  to 
five.  In  Problem  13,  the  forces  are  coplanar  in  the  first  position  and 
there  are  three  unknowns.  In  the  second  position,  the  problem 
involves  six  unknowns  In  Problem  14,  one  couple  is  left  out  and 
the  unknowns  are  reduced  to  five. 


CHAPTER  XII 
FRICTION 

112.  Coefficient  of  Friction. — Figure  171  represents  a  body  of  W 
pounds  mass  on  a  horizontal  plane.  A  rope  runs  horizontally 
from  the  body,  passes  over  a  smooth  pulley,  and  supports  a  mass 
of  P  pounds.  When  the  load  P  is  relatively  small,  the  system 
remains  stationary.  Regarding  the  mass  W  as  the  free  body 
and  resolving  horizontally,  the  tension  P'  in  the  horizontal  cord 
(which  is  nearly  equal  to  the  load  P)  is  balanced  by  the  horizontal 
force  from  the  supporting  plane  at 
the  surface  of  contact.  This  hori- 
zontal force  is  the  friction  between 
the  two  bodies  at  their  surface  of 
contact.  Friction  is  a  force  exerted 
between  two  bodies  at  their  surface  of 
contact.  The  force  of  friction  resists 
motion  of  one  body  parallel  to  the 

surface  of  the  other.  In  Fig.  171,  the  tension  P'  tends  to  pull 
the  mass  W  toward  the  right.  The  friction  from  the  plane  to  the 
body  acts  toward  the  left  as  is  indicated  by  the  arrow. 

If  the  load  P  is  gradually  increased,  the  tension  P'  will 
finally  reach  a  value  greater  than  the  friction.  The  mass  W  will 
begin  to  move  toward  the  right  and  continue  to  move  with  in- 
creasing speed.  If  an  additional  load  be  applied  to  the  mass  W, 
the  force  required  to  start  it  in  motion  will  also  be  increased.  The 
body  W  may  be  a  wooden  block  weighing  10  pounds  and  the  force 
required  to  start  it  may  be  4  pounds.  If  an  additional  10-pound 
weight  be  put  on  the  block,  the  force  required  to  start  it  will  be 
about  8  pounds. 

The  force  which  must  be  overcome  in  order  to  start  a  body  in 
motion  is  called  the  static  or  starting  friction. 

If  a  force  is  sufficient  to  start  a  body  in  motion,  that  force 
will  cause  the  body  to  move  with  increasing  speed  after  it  has 
been  started.  A  considerably  smaller  force  will  keep  it  in  motion 
with  uniform  speed,  after  it  has  been  started.  If  the  force  P' 
is  gradually  increased,  a  value  will  be  found  at  which  the  body 

189 


190  MECHANICS  [ART.  112 

will  continue  to  move  with  constant  speed  after  it  has  been  set 
in  motion  by  an  additional  force.  This  force  which  will  keep  the 
body  in  motion  with  constant  speed  after  it  has  been  started  is 
somewhat  less  than  the  force  required  to  start  it.  The  resistance 
when  the  body  is  moving  at  constant  speed  is  called  the  moving 
friction,  the  sliding  friction,  or  the  kinetic  friction. 

The  ratio  of  the  friction  to  the  pressure  normal  to  the  surface 
is  called  the  coefficient  of  friction. 

f-jpf-flfj  Formula  X 

in  which  /  is  the  coefficient  of  friction,  F  is  frictional  resistance, 
and  N  is  the  normal  pressure. 

Example 

In  Fig.  171,  the  mass  W  weighs  12  pounds  and  the  load  P  required  to 
start  it  in  motion  is  5.4  pounds.  The  ^ad  P  required  to  keep  the  body  in 
motion  after  it  has  been  started  is  4.2  pounds.  Find  the  coefficient  of  mov- 
ing friction  and  the  coefficient  of  starting  friction. 

If  the  friction  of  the  pulley  is  negligible,  it  may  be  assumed  that  the 
horizontal  tension  in  the  cord  is  equal  to  the  weight  P.  For  the  coefficient 
of  starting  friction, 

/.-M 

For  the  coefficient  of  moving  friction, 

"       '" 


The  coefficient  of  starting  friction  is  very  irregular.  For  moderate  loads, 
the  coefficient  of  moving  friction  between  surfaces  which  are  not  lubricated 
is  fairly  constant.  For  very  low  pressures,  and  for  very  high  pressures 
which  exceed  the  elastic  limit  of  the  material,  the  coefficient  of  moving 
friction  is  greater  than  it  is  for  moderate  pressures.  Friction  depends  upon 
the  material  and  upon  the  polish  of  the  surfaces  of  contact. 

Friction  is  greatly  reduced  by  lubrication  with  suitable  oils.  If  the  speed 
is  sufficient  to  keep  a  good  oil  film  between  the  surfaces,  the  friction  is  that 
of  solids  on  viscous  liquids,  and  the  coefficient  of  friction  decreases  with 
increase  of  pressure. 

In  the  problems  below  it  will  be  assumed  that  the  coefficient 
of  friction  is  constant. 

Problems 

1.  A  60-pound  mass  is  pulled  along  a  horizontal  plane  by  a  horizontal 
force  of  21  pounds.  Find  the  coefficient  of  friction.  Ans.  f  =  0.35. 

m 


CHAP.   XII] 


FRICTION 


191 


2.  A  60-pound  mass  is  pulled  along  a  horizontal  plane  by  a  force  of  20 
pounds  at  an  angle  of  20  degrees  above  the  horizontal.  Find  the  coefficient 
of  friction. 


Ans.    f  = 


18.79 
53.16 


0.353. 


FIG.  172. 


3.  The  coefficient  of  friction  between  a 
wooden  block  weighing  40  pounds  and  a  hori- 
zontal wooden  floor  is  0.32.    What  horizontal 
force  will  keep  the  block  moving  with  con- 
stant speed? 

4.  Solve  Problem  3  if  the  force  makes  an 
angle  of  20  degrees  above  the  horizontal. 

Resolving  horizontally,  Fig.  172, 

P  cos  20°  =  F  =  0.32  N. 
Resolving  vertically, 

N  =  40  -  P  sin  20°. 

Ans.    P  =  12.20  tb. 

6.  A  60-pound  mass  on  a  plane  inclined  25  degrees  to  the  horizontal  is 
just. pulled  up  the  plane  by  a  force  of  40  pounds  acting  upward  parallel  to 

14.64 

the  plane.     Find  the  coefficient  of  friction.         Ans.    f  =  — '-  -    =  0.27. 

54.38 

6.  In  Problem  5,  what  force  upward,  parallel  to  the  plane,  will  allow  the 
60-pound  mass  to  slide  down  the  plane  with  uniform  speed  after  it  has  once 
started?  Ans.     10.72  Ib. 

7.  In  Problem  5,  what  horizontal  force  is  required  to  keep  the  60-pound 
mass  moving  up  the  plane  with  uniform  speed?  Ans.     50.5  Ib. 

8.  A  ladder  24  feet  long  rests  on  a  horizontal  floor  and  leans  against  a 
smooth  vertical  wall.     The  ladder  makes  an  angle  of  20  degrees  with  the 
vertical.     The  center  of  gravity  of  the  ladder  and  its  load  is  16  feet  from 

the  bottom.  What  must  be  the  minimum 
value  of  the  coefficient"  of  friction  in  order 
that  the  ladder  may  not  slide  down  after  it 
has  been  started  by  a  slight  jar? 

Solve  by  one  moment  and  two  resolution 
equations,  together  with  Formula  X.  (Fig. 
173.)  Ans.  f  =  0.24. 

9.  Solve  Problem  8  if  the  wall  is  not  smooth 
but  has  a  coefficient  of  friction  equal  to  that 
of  the  horizontal  floor. 

10.  A  ladder  weighing  40  pounds,  with  its 
center  of  gravity  12  feet  from  the  lower  end, 
rests  on  a  horizontal  floor  and  leans  over  the 
upper  edge  of  a  smooth  vertical  wall  at  a 
distance   of  20  feet  from   the  floor.     The 

ladder  makes  an  angle  of  25  degrees  with  the  vertical.  The  coefficient  of 
friction  at  the  floor  is  0.35.  How  far  up  the  ladder  may  a  man  weighing  160 
pounds  climb?  Ans.  18.7  ft. 

11.  A  10-pound  mass  is  placed  on  a  30-degree  inclined  plane.     If  the 
coefficient  of  moving  friction  is  0.1,  what  force  parallel  to  the  plane  will 


•H 


FIG.  173. 


192  MECHANICS  [ART.  113 

just  pull  the  body  up  the  plane?     What  force  will  permit  the  body  to  slide 
down  with  uniform  speed?  Ans.     5.866  lb.;  4.13  Ib. 

12.  Solve  Problem  11  if  the  forces  are  horizontal.     Ans.  7.19  lb.;  4.52  lb. 

113.  Angle  of  Friction.  —  In  the  space  diagram  of  Fig.  174,  a 

body  of  W  pounds  mass  is  placed 
on  an  inclined  plane  which  makes 
an  angle  0  with  the  horizontal. 
The  component  of  the  weight 
down  the  plane  is  W  sin  <j>. 
The  component  normal  to  the 

Space  Diagram    Force  Triangle     plane  is  W  cos  0.     If  this  body 
FlG-  174>  slides  down  the  plane  with  uni- 

form speed  after  it  has  once  been  started,  the  component  of  its 
weight  parallel  to  the  plane  is  equal  to  the  moving  friction. 

F  =  W  sin  <£.  (1) 

=  tan  0.  (2) 


W  cos 

The  angle  at  which  a  plane  must  be  inclined  to  the  horizontal 
in  order  that  a  body  on  it  may  just  slide  down  with  uniform  speed 
is  called  the  angle  of  friction.  The  angle  at  which  a  plane  must 
be  inclined  in  order  that  a  body  may  start  down  the  plane  is 
called  the  angle  of  starting  friction.  If  the  inclination  of  a  plane 
slightly  exceeds  the  angle  of  sliding  friction,  a  slight  vibration 
will  set  the  body  in  motion  and  it  will  continue  to  slide  with 
increasing  velocity. 

Problems 

1.  In  Problem  1  of  Art.  112,  what  is  the  angle  of  friction? 

Ans.     0  =  19°  17'. 

2.  A  40-pound  mass  is  placed  on  a  plane  which  makes   an  angle  of  40 
degrees  with  the  horizontal.     It  is  found  that  a  pull  of  8  pounds  up  the 
plane  is  required  in  order  that  the  body  may  slide  down  the  plane  with 
uniform  speed.     Find  the  coefficient  of  friction  and  the  angle  of  friction. 

The  force  diagram  of  Fig.  174  shows  that  the  resultant  of  the 
force  of  friction  and  the  normal  force  is  a  vertical  force  which  is 
equal  and  opposite  to  the  weight  of  the  body.  If  the  inclination  of 
the  plane  were  less  than  the  angle  of  friction,  the  force  triangle 
would  be  similar  to  this  figure.  The  resultant  of  friction  and  the 
normal  reaction  would  be  vertical  and  equal  to  the  weight.  The 


CHAP.    XII] 


FRICTION 


193 


friction,  however,  would  be  less  than  its  limiting  value  and  the 
body  would  remain  stationary. 

In  Fig.  175,  four  forces  are  in  equilibrium.  The  resultant  of  F 
and  N  makes  an  angle  </>  with  the  normal  to  the  surface.  In  all 
cases  where  a  body  is  sliding  with  uniform  speed  there  is  a  force 
from  the  surface  of  con-  p 

A      I 

tact  to  the  moving  body, 
which    is  the   resultant  of 

the    normal    reaction    and         j-j-^ J  N  ^\  W 

the  moving  friction. 


W 


N 


B 


FIG.  175. 


The  angle  which  this 
resultant  force  makes  with 
the  direction  of  the  normal 
to  the  surface  is  the  angle  of  friction.  If  the  body  is  at  rest  at 
the  condition  of  incipient  sliding,  the  tangent  of  the  angle  which 
the  resultant  of  the  friction  and  the  normal  forces  makes  with 
the  normal  is  the  coefficient  of  starting  friction. 

Example 

A  body  is  moved  along  a  horizontal  plane  by  a  force  which  makes  an 
angle  6  with  the  horizontal.  What  is  the  value  of  0  in  order  that  the  force 
may  be  a  minimum? 

In  Fig.  176,  the  weight  W  is  laid  off  as  a  vertical  line  of  known  length. 
The  resultant  of  the  normal  and  the  friction  makes  an  angle  <£  with  the  ver- 
tical. The  tangent  of  <£  is  the  coefficient  of  friction.  Through  one  end  of 
the  vector  W  of  the  force  diagram,  a  line  of  indefinite  length  is  drawn  at 
an  angle  </>  with  the  vertical.  This  line  gives  the  direction  of  the  resultant 
of  the  friction  and  the  normal.  A  line  from  the  other  end  of  W  to  the  line 

of  this  resultant  represents  the  force  P. 
The  length  of  this  line  is  the  least 
when  it  is  perpendicular  to  the  line 
of  the  resultant.  Since  the  resultant 
makes  an  angle  0  with  the  vertical,  the 
line  P  perpendicular  to  the  resultant 
makes  the  same  angle  with  the  hori- 
zontal. The  pull  is  a  minimum  when 
its  angle  with  the  horizontal  plane  is 
the  angle  of  friction. 


W 


W 


Tan<t>»f 


Force  Diagram 


FIG.  1761 


Problems 

3.  A  body  is  pulled  up  an  inclined  plane  by  a  force  which  makes  an  angle 
6  with  the  plane.     Show  graphically  that  the  pull  is  the  least  when  its  angle 
with  the  plane  is  equal  to  the  angle  of  friction. 

4.  Solve  Problem  8  of  the  preceding  article  by  means  of  the  direction 
condition  of  equilibrium. 

13 


194  MECHANICS  [ART.   114. 

6.  A  body  is  drawn  along  a  horizontal  plane  by  a  force  which  makes  an 
angle  with  the  plane  equal  to  the  angle  of  friction.  Show  that  this  force  is 
the  product  of  the  weight  of  the  body  multiplied  by  the  sine  of  the  angle  of 
friction. 

When  a  body  is  stationary  or  moving  with  uniform  speed  along 
a  surface,  the  resultant  of  the  friction  and  the  normal  reaction 
of  the  surface  is  equal  and  opposite  to  the  resultant  of  all  the 
other  forces  which  act  on  the  body.  If  the  body  is  moving  with 
uniform  speed,  the  resultant  of  the  normal  and  the  friction  makes 
an  angle  with  the  normal  whose  tangent  is  the  coefficient  of 


A\ 

\ 
\ 

\ 

P 


NB 

FIG.   177. 

sliding  friction.     The  resultant  of  all  the  other  forces  makes  the 
same  angle  with  the  normal.     In  Fig.  177,  the  external  forces  are 
W  and  P.     Their  resultant  is  R.     This  resultant  balances  the 
resultant  of  the  friction  and  the  normal. 
If  <£  is  the  angle  of  friction, 

tan  </>  =  TF;  (3) 

sin0     =J  (4) 

In  the  case  of  lubricated  surfaces,  where  the  coefficient  of  friction 
is  small,  tan  <£  is  practically  equal  to  sin  0,  and  the  resultant  R 
differs  little  from  the  normal  N. 

114.  Cone  of  Friction. — Figure  178,  I,  shows  a  body  pushed 
along  a  plane  by  a  force  which  has  a  large  component  normal  to 
the  plane.  The  component  against  the  plane  greatly  increases  the 
normal  pressure  and,  as  a  result,  greatly  increases  the  friction. 
Figure  178,  II,  is  the  force  diagram.  If  the  angle  between  the 
force  P  and  the  plane  is  further  increased,  the  vectors  P  and  R 
of  Fig.  178,  II,  will  become  more  nearly  parallel,  will  intersect  at 


CHAP.    XII] 


FRICTION 


195 


greater   distances,   and   will   represent   correspondingly  greater 
forces. 

In  Fig.  178,  III,  the  force  P  makes  an  angle  with  the  vertical 
which  is  equal  to  the  angle  of  friction.  The  force  polygon  will 
not  close.  It  is  impossible  to  move  the  body  by  a  force  in  this 
direction.  If  the  force  makes  a  still  smaller  angle  with  the 


FIG.   178. 

vertical,  as  shown  in  Fig.  178,  IV,  the  force  polygon  will  not  close 
when  the  force  is  downward  toward  the  surface.  When  a  force 
which  pushes  a  body  against  a  surface  makes  an  angle  with  the 
normal  which  is  less  than  the  angle  of  friction,  this  force  will 
not  move  the  body  along  the  surface,  no  matter  how  great  it  may 
be.  The  lines  which  make  the  same  angle  with  the  normal  to 
a  surface  form  a  cone,  as  shown  in  Fig.  179.  When  this  angle 
is  the  angle  of  friction,  the  cone  is  called  the  cone  of  friction.  A 
force  which  is  applied  along  a  direction  inside 
the  cone  of  friction  for  a  body  and  surface  has 
no  tendency  to  move  the  body  along  the  sur- 
face. The  component  of  such  a  force  normal 
to  the  surface  increases  the  friction  more  than 
the  tangential  component  increases  the  com- 
ponent parallel  to  the  surface. 

When  a  body  is  moved  along  a  surface,  the 
applied  force  must  be  outside  the  cone  of  fric- 
tion for  the  body  and  the  surface.     A  force,  applied  inside  the 
direction  of  the  cone  of  friction,  tends  to  hold  the  body  stationary. 

115.  Bearing  Friction. — Figure  180  shows  a  shaft  of  radius  a 
turning  in  a  very  loose  bearing  in  a  clockwise  direction.  No 
such  bearing  would  be  used  in  practice.  It  is  drawn  in  this  way 


FIG.  179. 


196 


MECHANICS 


[ART.    115 


in  Fig.  180  in  order  to  show  details  more  clearly.  When  the 
shaft  is  turned,  it  climbs  up  on  the  bearing  until  it  reaches  a 
point  at  which  it  slides  down  as  fast  as  it  rolls  up.  At  this 
point,  the  resultant  pressure  makes  an  angle  with  the  normal 
which  is  equal  to  the  angle  of  friction  for  the  surfaces  of  the  shaft 
and  the  bearing. 

If  there  is  a  wheel  rigidly  attached  to  the  shaft,  and  if  forces  P 
and  Q  are  applied  to  this  wheel  by  means  of  a  belt,  the  resultant 
moment  of  these  forces  about  the  axis  of  the  shaft  must  be  equal 
to  the  moment  of  the  friction  about  that  axis.  In  Fig.  180,  the 


FIG.  180. 

moment  of  the  friction  is  counter-clockwise.  The  moment  of 
the  force  P  must  then  be  greater  than  the  moment  of  the  force 
Q.  The  line  of  the  normal  at  the  point  of  contact  B  passes 
through  the  axis  of  the  shaft.  The  resultant  of  the  friction  and 
the  normal  reaction  of  the  bearing  makes  an  angle  0  with  the 
direction  of  this  normal.  The  resultant  of  the  external  forces  P 
and  Q  passes  through  their  intersection  at  C.  If  the  weight  of 
the  wheel  and  of  the  shaft  are  neglected,  the  resultant  of  the 
forces  P  and  Q  will  pass  through  B.  The  resultant  of  P  and  Q 
will  be  equal  and  opposite  to  the  resultant  of  the  normal  reaction 
and  the  friction  at  B,  and  will  lie  along  the  same  line. 

If  a  circle  of  radius  a  sin  0  is  drawn  concentric  with  the  axis 
of  the  shaft,  the  line  of  the  resultant  reaction  at  B  will  be  tangent 
to  this  circle.  This  circle,  called  the  friction  circle,  is  convenient 
for  some  graphical  solutions. 


CHAP.   XII] 


FRICTION 
Example 


197 


In  Fig.  181  the  shaft  is  6  inches  in  diameter.  The  wheel  is  2  feet  in  diam- 
eter and  weighs  200  pounds.  The  force  P  is  240  pounds.  The  force  Q  is 
200  pounds.  Both  forces  are  vertically  downward.  Find  the  coefficient 
of  friction. 

Taking  moments  about  the  axis  of  the  shaft, 

3F  =  240  X  12  -  200  X  12  =  (240  -  200)  X  12  =  480  ft.-lb. 
F  =  160  Ib. 

Since  all  the  applied  forces  are  parallel,  their  resultant  is  equal  to  their  sum. 
The  resultant  is  640  pounds  vertically  downward. 


14°  29'. 

tan  <f>  =  0.258. 


FIG.  181. 

The  student  will  understand  that  this  bearing  is  not  lubricated.  The 
coefficient  of  friction  for  a  well  lubricated  bearing  should  be  many  times 
smaller  than  0.258. 

Since  all  the  applied  forces  are  vertical,  this  problem  may  be  conveniently 
solved  by  moments  about  the  point  of  contact  of  the  shaft  and  bearing. 
The  equation  of  moments  about  this  point  B  of  Fig.  181  is 

200(12  +  3  sin  0)  +  200  X  3  sin  <£  =  240(12  -  3  sin  0) 

Problems 

1.  In  the  Example  above,  the  force  Q  is  vertical  and  the  force  P  is  hori- 
zontal toward  the  right.  Find  the  coefficient  of  friction,  and  the  position 
of  the  point  of  contact  of  the  shaft  and  bearing. 


198  MECHANICS  [ART.   116 

Ans.  R  =  467  lb.;  /  =  0.365.  The  normal  at  the  point  of  contact 
makes  an  angle  of  20°  04'  +  30°  58'  =  51°  02'  with  the  vertical. 

2.  An  axle  4  inches  in  diameter  is  connected  to  a  wheel  2  feet  in  diameter. 
The  wheel  and  axle  together  weigh  160  pounds.     A  rope  passes  over  the 
wheel  and  hangs  vertically  downward  on  each  side.     A  load  of  240  pounds 
is  placed  on  one  end  of  the  rope.     The  coefficient  of  friction  is  0.16.     What 
is  the  load  on  the  other  end  of  the  rope  which  will  lift  the  240-pound  load? 

Ans.     257.3  lb. 

3.  Solve  Problem  2  if  the  axle  is  only  2  inches  in  diameter  at  the  bearings. 

4.  Solve  Problem  3  if  the  bearings  are  so  well  lubricated  that  the  coeffi- 
cient of  friction  is  0.03. 

Approximate  solutions  of  problems  of  bearing  friction  are 
often  sufficiently  accurate.  One  approximation  is  the  assump- 
tion that  the  resultant  is  equal  to  the  normal  pressure.  This 
is  equivalent  to  R  =  N,  and  sin  0  =  tan  0.  When  the  coefficient 
of  friction  is  small  this  method  involves  little  relative  error. 
Another  approximation  is  the  calculation  of  the  resultant  (but 
not  the  moment)  on  the  assumption  that  P  =  Q.  In  Problem 
2,  if  it  is  assumed  that  P  =  Q  in  the  calculation  of  the  resultant, 

R  =  240  +  240  +  160  =  640  lb. 
If  it  is  assumed  that  R  =  N, 

F  =  0.16  X  640  =  102.40  lb. 
Taking  moments  about  the  axis  of  the  shaft, 

(Q  -  240)  X  12  =  102.40  X  2; 

(Q  -  240)  =  17.07  lb; 

Q  =  257.07  lb. 

It  is  not  probable  that  the  coefficient  of  friction  is  known  with 
anything  like  the  relative  accuracy  of  this  result. 

Problems 

6.  Solve  Problem  3  by  the  approximate  methods. 

6.  An  axle  4  inches  in  diameter  is  attached  to  a  wheel  4  feet  in  diameter. 
The  wheel  and  axle  together  weigh  200  pounds.  A  rope  hangs  vertically 
downward  on  one  side  of  the  wheel  and  supports  a  load  of  300  pounds. 
On  the  other  side  of  the  wheel  the  rope  makes  an  angle  of  30  degrees  below 
the  horizontal.  What  is  the  tension  in  this  part  of  the  rope  required  to 
lift  the  load  of  300  pounds  if  the  coefficient  of  friction  is  0.12?  Solve  by 
the  approximate  method. 

116.  Rolling  Friction. — When  a  cylinder  rolls  on  a  surface,  the 
surface  is  depressed  somewhat  by  the  pressure  which  the  cylinder 
exerts  on  it.  This  is  shown  on  an  exaggerated  scale  in  Fig.  182. 


CHAP.  Xll]  FRICTION  199 

If  the  material  of  the  surface  is  resilient,  it  will  spring  back  so  as 
to  push  against  the  cylinder  on  both  sides  of  the  lowest  point. 
If  its  resilience  is  small,  it  will  come  back  slowly  so  that  practi- 
cally all  the  pressure  will  be  in  front  of  the  lowest  point  of  the 
cylinder.  In  any  case,  the  reaction  of  that  part  of  the  surface  in 
front  of  the  cylinder  will  be  greater  than  the  reaction  on  the 
other  side,  and  the  resultant  of  all  the  reactions  will  be  somewhat 
as  shown  in  the  figures.  The  cylinder  may  be  regarded  as  con- 
tinuously climbing  a  small  hill.  Experiments  have  shown  that 
the  distance  of  the  resultant  reaction  from  the  line  of  the  load  W 
is  practically  constant  for  a  given  material  and  is  independent 
of  the  diameter  of  the  cylinder.  This  distance,  fr  of  Fig.  182, 
is  the  coefficient  of  rolling  friction.  For  steel  on  steel  the  coeffi- 
cient of  rolling  friction  is  from  0.02  inch  to  0.03  inch. 


FIG.  182. 

In  Fig.  182,  I,  the  cylinder  is  rolled  along  a  horizontal  surface 
by  a  horizontal  force  applied  at  the  center.  If  moments  are 
taken  about  the  point  of  contact  B,  the  moment  arm  of  the  force 
P  is  practically  equal  to  the  radius  of  the  cylinder.  The  moment 
arm  of  the  weight  W  is  the  coefficient  of  rolling  friction  fr.  The 
moment  equation  is  then, 

Wfr  =  Pa.  (1) 

In  Fig.  182,  II,  the  cylinder  is  rolled  along  by  means  of  a  force 
parallel  to  the  surface  applied  at  the  top.  The  moment  arm  of 
this  force  is  the  diameter  of  the  cylinder. 

Wfr  =  2Pa.  (2) 

Problems 

1.  If  the  coefficient  of  rolling  friction  of  iron  on  iron  is  0.03  inch,  what 
horizontal  force  applied  at  the  center  of  an  iron  wheel,  18  inches  in  diameter, 
weighing  200  pounds,  will  just  roll  it  along  a  smooth,  horizontal  iron  surface. 

Ans.     0.67  Ib. 


200  MECHANICS  [ART.  117 

2.  An  iron  wheel,  30  inches  in  diameter,  weighing  160  pounds,  rolls  on  a 
steel  rail.  The  coefficient  of  rolling  friction  is  0.03  inch.  The  wheel  is 
attached  to  an  axle  4  inches  in  diameter.  A  load  of  800  pounds  is  applied 
to  the  axle  by  means  of  a  bearing.  The  coefficient  of  sliding  friction  between 
the  axle  and  the  bearing  is  0.04.  What  horizontal  pull  is  required  to  move 
the  load?  Ans.  6.32  Ib. 

3.  Figure  183  represents  a  mass  of  W  pounds 


]\   '  W  ^y  /'\    >  P    on  cylindrical  rollers.     The  coefficient  of  roll- 

?Hr ' — /<  \  ing  friction  is  0.04  inch  at  the  bottom  and  0.05 

wyiw^JfLw*-^^  i110*1  at  *he  toP-     ^e  cylinders  are  10  inches 

FIG.  183.  *n  diameter  and  weigh  100  pounds  each.    The 

load  W  is  2000  pounds.     If  the  track  is  hori- 
zontal, what  is  the  force  required  to  keep  the  load  moving? 

Ans.     18.8  Ib. 

4.  In  Problem  3,  what  is  the  pull  required  to  keep  the  load  moving  up  a 
1-degree  inclined  plane? 


117.  Roller  Bearings. — Since  the  coefficient  of  rolling  friction 
of  hard  steel  on  hard  steel  is  very  small,  it  is  desirable  to  design 
bearings  in  which  the  friction  is  rolling  rather  than  sliding.     This 
is  accomplished  by  means  of  roller  bearings,   or  ball  bearings. 
Fig.  184  shows  a  roller  bearing.     The  axle  rotates  on  hardened 
steel  rollers  which  in  turn  roll  on  the  inside  surface  of  a  hollow 
steel  cylinder.     In  order  to  keep  the  rollers  properly  spaced  so 
that  they  will  not  come  together  and  introduce  sliding  friction, 
each  roller  is  connected  at  the  ends  to  a  light  metal  frame  called 
the  cage. 

For  relatively  light  loads,  ball  bear-  ^^} 

ings  are  employed  instead  of  roller 
bearings.  Spherical  balls  roll  in 
grooves  or  races,  which  may  be  V 
shaped  or  circular. 

For  rolling,  the  coefficient  of  start- 
ing friction  is  nearly  the  same  as  the 
coefficient  of  moving  friction.  This 
property  is  one  of  the  elements  of  the 
superiority  of  roller  or  ball  bearings  FIG.  184. 

over  the  ordinary  bearings. 

118.  Belt  Friction. — Figure  185,  I,  represents  a  pulley  and  a 
belt.     The  belt  is  in  contact  with  the  pulley  through  an  angle  of 
a  radians.     The  tension  of  the  belt  at  the  left  point  of  tangency 
with  the  pulley  is  PI.     The  tension  at  the  right  point  of  tangency 


CHAP.   XII] 


FRICTION 


201 


is  P2.     The  tension  P2  is  greater  than  PI  so  that  the  belt  tends 
to  turn  the  pulley  in  a  clockwise  direction. 

P2  —  P\  =  friction  between  belt  and  pulley. 

Figure  185,  II,  shows  a  small  element  of  the  length  of  the  belt. 
This  element  is  enclosed  between  radii  which  make  an  angle  dB 
with  each  other.  The  tension  on  the  left  end  of  the  element  is  P; 
on  the  right  end  is  P  -f  dP.  The  difference  between  these  two 
tensions  is  dP.  This  difference  is  equal  to  the  friction  between 
the  belt  and  the  pulley.  Since  the  two  tensions,  P  and  P  +  dP, 


FIG.  185. 

are  each  tangent  to  the  pulley,  they  make  an  angle  d0  with  each 
other,  and  each  tension  makes  an  angle  -~   with  the  tangent  at 

the  middle  of  the  element.     The  normal  reaction  of  the  pulley  on 
the  element  of  the  belt  is  obtained  by  a  resolution  along  the 

radius  BO.     The  component  of  P  is  P  sin  -^>  which,  for  a  small 

d6 
angle,  is  equal  to  P  -^-     When  the  differentials  of  the  second 

z 

order  are  omitted,  the  component  of  P  +  dP  is  also  P  -*-     The 
total  normal  force  on  the  element  is 

N  =  Pdd.  (1) 

If  the  coefficient  of  friction  between  the  belt  and  the  pulley  is  /, 
then 


F  =  fN  =  fPdS  =  dP. 


(2) 


202  MECHANICS  [ART.  118 

Separating  the  variables  in  this  last  equation, 

^  =  fdO;  (3) 

lo&P  =  fB  +  C.  (4) 

When  0  =  0,  P  =  PI;  C  =  log(Pi 

/0    =    log,P    -    logePi.  (5) 

When  0  =  a,P  =  P2; 

...     (6) 


Example 

A  belt  is  in  contact  with  a  pulley  through  an  angle  of  180  degrees.  The 
tension  at  one  point  of  tangency  is  100  pounds  and  the  coefficient  of  friction 
is  0.4.  What  is  the  maximum  tension  at  the  other  point  of  tangency? 

loge  ^  =  3.1416  X  0.4  =  1.2566. 
/i 

If  a  suitable  table  of  Naperian  logarithms  were  available,  the  ratio  of  P2 
to  PI  could  be  found  by  looking  up  the  number  whose  logarithm  is  1.2566. 
Without  such  a  table,  the  Naperian  logarithms  must  be  reduced  to  common 
logarithms. 

logic         =  0.5457, 


P2  =  100  X  3.51  =  351  Ib. 

The  tension  of  351  pounds  is  the  greatest  value  possible  without  slipping. 
The  actual  tension  may  be  very  much  less  than  351  pounds. 

Problems 

1.  A  rope  is  wound  twice  around  a  cylindrical  post.     The  coefficient  of 
friction  is  0.3.     What  must  be  the  minimum  tension  in  one  end  of  the  rope 
in  order  to  have  a  tension  of  1,000  pounds  in  the  other  end? 

Ans.     23.05  Ib. 

2.  A  rope  is  hung  over  a  cylinder  whose  axis  is  horizontal.     When  a 
load  of  20  pounds  is  hung  on  one  end  of  the  rope,  it  is  found  that  a  load  of 
60  pounds  may  be  hung  on  the  other  end.     If  the  rope  were  given  an  addi- 
tional turn  around  the  cylinder,  how  many  pounds  could  be  hung  on  one 
end  when  there  is  a  load  of  20  pounds  on  the  other  end?     Solve  without 
writing. 

3.  A  cylindrical  drum  10  inches  in  diameter  is  used  to  lift  a  load  of  2,000 
pounds  by  means  of  a  1-inch  rope.     The  drum  is  turned  by  means  of  a 
pulley  4  feet  in  diameter  which  is  attached  at  one  end.     The  pulley  is  driven 


CHAP.   XII]  FRICTION  203 

by  a  belt  from  a  second  pulley  of  the  same  diameter.  If  the  coefficient  of 
friction  between  the  belt  and  the  pulley  is  0.4,  what  must  be  the  tension  in 
the  two  parts  of  the  belt?  Ans.  641  lb.;  183  Ib. 

4.  Solve  Problem  3  if  the  second  pulley  is  1  foot  in  diameter  and  its  axis 
is  5  feet  from  the  axis  of  the  drum. 

119.  Summary. — Friction  is  a  force  at  the  surface  of  contact  of 
two  bodies,  which  resists  the  motion  of  one  body  along  the  surface 
of  the  other.  The  force  of  friction  is  tangent  to  the  common 
surface  of  the  two  bodies  and  is  opposite  the  direction  which  one 
body  is  moving  or  tending  to  move  relative  to  the  other  body. 

The  ratio  of  the  friction  to  the  normal  pressure  is  called  the 
coefficient  of  friction.  The  coefficient  of  moving  friction  is 
the  value  of  this  ratio  when  the  bodies  actually  move  relative  to 
each  other.  The  coefficient  of  starting  friction  is  the  maximum 
value  of  this  ratio  when  there  is  no  motion.  The  coefficient  of 
starting  friction  is  greater  than  the  coefficient  of  moving  friction. 

For  solids  which  are  not  lubricated,  the  coefficient  of  friction 
is  fairly  constant  with  varying  pressure  and  velocity.  The 
coefficient  varies  greatly  with  the  smoothness  of  the  surface  and 
the  kind  of  material. 

MATERIAL  COEFFICIENT  OF  MOVING  FRICTION 

Wood  on  wood,  dry 0 . 25  to  0 . 50 

Metal  on  metal,  dry 0 . 15  to  0 . 20 

Leather  on  metal,  dry 0 . 35  to  0 . 55 

When  the  surfaces  are  well  lubricated,  the  coefficient  of  friction 
decreases  with  increase  of  pressure  and  is  much  smaller  than  for 
dry  surfaces. 

The  ratio  of  the  friction  to  the  normal  pressure  (the  coefficient 
of  friction)  is  the  tangent  of  the  angle  of  friction.  The  ratio  of  the 
friction  to  the  resultant  pressure  is  the  sine  of  the  angle  of  friction. 

If  <f>  is  the  angle  of  friction,  the  resultant  of  all  the  applied 
forces,  except  the  normal  reaction  and  the  friction,  makes  an 
angle  <f>  with  the  direction  of  the  normal. 

The  coefficient  of  rolling  friction  is  the  distance  of  the  point  of 
application  of  the  resultant  reaction  at  the  surface  of  the  rolling 
body  from  a  line  through  the  center  of  the  rolling  body  parallel 
to  the  resultant  of  all  loads  except  the  force  required  to  roll 
the  body. 

The  coefficient  of  rolling  friction  is  practically  independent 
of  the  radius  of  the  rolling  body.  The  force  required  to  roll 
a  body  varies  inversely  as  its  diameter. 


204  MECHANICS  [ART.   120 

In  the  case  of  a  belt  running  over  a  pulley, 


loglo^-2  =  0.4343  fa, 
r\ 

in  which  P2  is  the  tension  in  the  belt  at  one  point  of  tangency, 
PI  is  the  tension  at  the  other  point  of  tangency,  /  is  the  coefficient 
of  friction,  and  a  is  the  angle  of  contact  in  radians. 

120.  Miscellaneous  Problems 

1.  A  cylinder,  8  inches  in  diameter,  weighing  6  pounds,  is  placed  on  two 
parallel  rods,  which  make  an  angle  of  15  degrees  with  the  horizontal.     A 
string  is  wound  around  the  cylinder  and  supports  a  weight  on  the  free  end. 
What  must  be  the  weight  in  order  to  hold  the  cylinder  in  equilibrium,  and 
what  must  be  the  minimum  value  of  the  coefficient  of  friction?     What  is  the 
direction  of  the  resultant  of  the  friction  and  the  normal  at  the  points  of 
contact  with  the  rods? 

2.  A  cylinder,  8  inches  in  diameter,  and  24  inches  in  length,  weighs  18 
pounds  and  has  its  center  of  gravity  at  the  middle.     The  cylinder  is  sup- 
ported by  a  rod  perpendicular  to  its  length  at  a  distance  of  9  inches  from 
one  end,  and  by  a  cord  which  is  wound  around  it  and  supported  above.     If 
the  coefficient  of  friction  is  sufficient  to  prevent  sliding,  how  far  must  the 
cord  be  placed  from  the  other  end  of  the  cylinder?     Solve  by  moments. 
Also  solve  graphically. 

3.  A  car  is  56  inches  wide,  center  to  center  of  wheels,  and  108  inches 
long,  center  to  center  of  axles.     The  center  of  gravity  is  30  inches  above 
the  ground  and  60  inches  in  front  of  the  rear  axle.     If  the  coefficient  of 
starting  friction  is  0.30  and  the  brakes  are  set  so  that  the  wheels  will  turn, 
what  is  the  steepest  grade  which  the  car  can  descend  without  accelerating? 

Ans.     7°  01'. 

4.  What  is  the  steepest  grade  which  it  is  possible  for  the  car  of  Problem  3 
to  ascend?  Ans.     8°  16'. 


CHAPTER  XIII 
WORK  AND  MACHINES 

121.  Work  and  Energy. — In  Art.  39,  the  work  done  by  a  force 
was  defined  as  the  product  of  the  displacement  of  its  point  of 
application  multiplied  by  the  component  of  the  force  in  the  direc- 
tion of  the  displacement.  Expressed  algebraically, 

U  =  P  cos  a  s,  Formula  III 

in  which  P  is  the  force;  s  is  the  displacement  or  amount  of  motion 
of  its  point  of  application;  a  is  the  angle  between  the  direction 
of  the  force  and  the  direction  of  the  displacement;  and  U  is  the 
work.  Since 

P  cos  a  X  s  =  P  X  s  cos  a,  (1) 

work  may  be  defined  as  the  product  of  the  entire  force  multiplied 
by  the  component  of  the  displacement  in  the  direction  of  the 
force. 

Work  is  the  scalar  product  of  two  vectors,  and  is  not  a  vector. 
Quantities  of  work  may  be  added  as  mere  numbers  with  no  regard 
to  the  directions  of  the  forces  and  displacements. 

Example  I 

A  10-pound  mass  is  lifted  4  feet  vertically  upward.  Find  the  work  done 
in  lifting  it. 

The  average  force  is  10  pounds  vertically  upward.  If  the  body  was 
initially  at  rest,  a  force  a  little  greater  than  10  pounds  was  required  to  set 
it  in  motion.  As  it  was  brought  to  rest  at  the  end  of  the  displacement, 
the  force  was  a  little  less  than  10  pounds.  The  point  of  application  of 
the  force  moves  in  the  direction  of  the  force  so  that  cos  a  =  1. 

U  =  10  X  4  =  40  foot-pounds. 

Example  II 

A  10-pound  mass  is  pulled  12  feet  up  a  30-degree  smooth  inclined  plane 
by  a  force  which  makes  an  angle  of  15  degrees  with  the  plane.  Find  the 
work  done. 

A  resolution  parallel  to  the  plane  gives, 

P  cos  15°  =  10  sin  30°  =  5  lb.; 
U  =  5  X  12  =  60  foot-pounds. 

Instead  of  taking  the  body  up  the  inclined  plane  of  Fig.  186,  it  might 
have  been  taken  horizontally  from  A  to  B,  and  then  lifted  vertically  from 

205 


206  MECHANICS  [ART.  121 

B  to  C.  If  there  is  no  friction,  no  work  is  done  in  taking  the  body  from 
A  to  B  (since  the  force  is  normal  to  the  displacement).  A  force  of  10 
pounds  is  required  to  lift  the  body  vertically  upward  from  B  to  C.  The 
displacement  from  B  to  C  is  6  feet  and  the  work  is  60  foot-pounds. 

When  the  force  varies  in  mag- 
nitude or  direction,   or  when  the 

'°*\     /fe^^n  displacement   varies    in   direction, 

the  total  work  is  calculated  from 
the  work  of  a  large  number  of 
small  displacements.  If  ds  is  an 
element  of  the  displacement  so 
small  that  the  magnitude  and 

direction  of  the  force  and  the  direction  of  the  displacement  may 
be  taken  as  constant  throughout  that  interval,  the  increment  of 
work  is  given  by  the  equation, 

dU  =  Pcosads.  (2) 

The  total  work  is  the  integral  of  Equation  (2), 

U  =  fP  cos  a  ds.  (3) 

Problems 

1.  A  90-pound  mass  on  a  horizontal  plane  is  moved  60  feet  north  by  a 
force  of  50  pounds  which  is  directed  north  20  degrees  east.     Find  the  work 
done  by  this  force.  Ans.     2819  ft.-lb. 

2.  It  requires  a  force  of  12  pounds  to  stretch  a  given  spring  a  distance  of 
1  foot.     If  the  force  is  proportional  to  the  elongation,  what  is  the  work  done 
in  stretching  the  spring  1  foot?     Solve  by  means  of  the  average  force  and 
check  by  integrating  Equation  3.  Ans.     6  ft.-lb. 

3.  After  the  spring  of  Problem  2  has  been  stretched  1  foot,  find  the  work 
required  to  stretch  it  an  additional  foot.  Ans.     18  ft.-lb. 

When  a  body  is  lifted  vertically  upward,  work  is  done  on  it. 
When  the  body  comes  down  to  its  original  position,  it  may  do  an 
equal  amount  of  work  on  another  body.  When  a  spring  is 
stretched,  work  is  stored  up  in  it  and  this  work  may  later  be 
given  to  another  body.  Work  which  is  stored  up  in  a  body  is 
called  energy.  The  energy  of  a  deformed  elastic  body,  or  the 
energy  of  a  body  which  has  been  lifted  against  the  force  of  gravity, 
is  called  potential  energy.  Potential  energy  is  sometimes  defined 
as  the  energy  of  position.  The  energy  of  a  moving  body  is  called 
kinetic  energy ,  or  the  energy  of  motion.  In  Example  1,  the  potential 
energy  of  the  10-pound  mass,  after  it  has  been  lifted  4  feet,  is 
40  foot-pounds  more  than  the  energy  before  it  was  lifted.  This 
gravitational  potential  energy  may  be  regarded  as  the  energy 


CHAP.  XIII]  WORK  AND  MACHINES  207 

of  deformed  ether.  In  Problem  2,  the  potential  energy  of  the 
spring  is  6  foot-pounds  when  it  is  stretched  1  foot,  and  is  24  foot- 
pounds when  it  is  stretched  2  feet. 

Kinetic  energy  will  be  discussed  in  Chapter  XVI. 

122.  Equilibrium  by  Work. — When  a  body  which  is  in  equi- 
librium under  the  action  of  several  forces  is  displaced,  the  total 
work  done  by  all  the  forces  is  equal  to  zero.     This  is  true  because, 
when  a  body  is  in  equilibrium,  the  resultant  force  is  zero  and  the 
component  of  the  resultant  along  any  direction  of  displacement 
is  zero.     This  principle  affords  a  convenient  method  of  solving 
some  problems  of  equilibrium.     The  method  is  especially  valu- 
able when  the  body  is  constrained  to  move  along  some  smooth 
surface.     In  the  application  of  the  method,  the  body  is  assumed 
to  suffer  a  displacement  along  the  surface,  and  the  total  work  of 
all  the  forces  is  equated  to  zero.     If  the  direction  and  magnitude 
of  the  forces  remain  constant  through  a  large  displacement,  the 
assumed  displacement  may  be  taken  as  equally  large.     If  either 
the  direction  or  the  magnitude  of  any  force  changes  with  a  small 
displacement,  the  assumed  displacement  is  taken  as  an  infinitesi- 
mal length. 

123.  Machines. — A  body  or  combination  of  bodies  for  the 
transfer  or  transformation  of  work  is  called  a  machine.     The 
input  is  the  work  done  on  a  machine.     The  output  is  the  useful 
work  done  by  the  machine.     The  ratio  of  the  useful  work  done 
by  the  machine  to  the  input  is  the  efficiency  of  the  machine. 
Suppose  that  a  rope  over  a  single  pulley  is  employed  to  lift  a 
load  of  40  pounds  vertically  upward,  and  suppose  that  the  pull 
on  the  rope  on  the  other  side  of  the  pulley  is  50  pounds.     When 
the  load  is  lifted  1  foot,  the  useful  work  is  40  foot-pounds.     If 
the  rope  does  not  stretch,  the  input  is  50  foot-pounds.     The 
efficiency  is  80  per  cent.     The  10  foot-pounds  of  work  are  lost 
in  friction  of  the  pulley,  and  are  transformed  into  heat.     If  there 
were  no  friction  or  similar  losses,  the  efficiency  would  be  one 
hundred  per  cent. 

The  force  which  acts  on  a  machine  with  a  component  in  the 
direction  of  the  displacement,  and  which  does  positive  work  on 
the  machine,  is  called  the  effort.  The  force  which  does  negative 
work  on  the  machine,  or  with  which  the  machine  does  work,  is 
called  the  resistance.  The  ratio  of  the  resistance  to  the  effort  is 
called  the  mechanical  advantage  of  the  machine.  When  a  force 
of  20  pounds  on  one  arm  of  a  lever  lifts  a  mass  of  60  pounds 


208 


MECHANICS 


[ART.    124 


w 


vertically  upward,  the  mechanical  advantage  of  that  lever  as  a 
machine  is  ^  =  3.  Since  the  work  done  by  the  effort  is  equal  to 

the  negative  work  of  the  resistance,  (assuming  no  friction),  the 
mechanical  advantage  is  equal  to  the  ratio  of  the  displacement 
of  the  effort  to  the  displacement  of  the  resistance. 

124.  Inclined  Plane. — In  Fig.  187,  a  body  on  an  inclined  plane 
is  supported  by  a  force  parallel  to  the  plane. 
The  supporting  force  P  and  the  mechanical 
advantage  will  be  found  by  the  method  of 
work.  If  the  length  of  the  plane  is  I,  the 
work  done  by  the  force  P  in  pulling  the  body 
that  distance  is  PL  If  the  height  of  the  plane 
is  h,  the  negative  work  of  gravity  on  the  body  of  W  pounds  mass 
is  Wh.  The  normal  reaction  of  the  plane  does  no  work,  since 
its  direction  is  perpendicular  to  the  displacement.  If  there  is 
no  friction,  the  input  is  equal  to  the  output,  and 

PI  =  Wh,  (1) 

Wh 


FIG.  187. 


P  = 


=  W  sin  0 


Equation  (2)  is  the  same  as  is  obtained  by  a  resolution  parallel 
to  the  inclined  plane. 


Equation  (3)  states  that  the  mechanical  advantage  of  an  inclined 
plane,  when  the  force  is  applied  parallel  to  the  plane,  is  equal  to 
the  ratio  of  the  length  of  the  plane  to  its 
height,  or  is  equal  to  the  cosecant  of  the  wl       ^^^ 

angle  of  slope  of  the  plane.  JpwX\        l<: 

In  Fig.  188,  the  force  P,  which  acts  on    ^^      p 
the  body,  is  horizontal.     When  the  body 
moves  a  distance  /  along  the  plane,  the  com- 
ponent of  its  displacement  in  the  direction 
of  the  force  is  the  horizontal  projection  of  that  distance.     The 
work  done  by  the  force  P  is  Pb,  in  which  b  is  the  horizontal 
projection  of  I. 

Pb  =  Wh,  (4) 

p  =  I™  =  W  tan  0.  (5) 


FIG.  188. 


W      b 

-^  =  7  =  cotan  6. 

r        n 


(6) 


CHAP.  XIII]  WORK  AND  MACHINES  209 

Equation  (5)  states  that  the  mechanical  advantage  of  an  inclined 
plane,  when  the  force  is  horizontal,  is  equal  to  the  ratio  of  the 
base  of  the  plane  to  its  height. 

When  a  body  is  lifted  up  an  inclined  plane,  the  output  is  in 
the  form  of  the  increased  potential  energy  at  the  new  position. 

In  any  machine,  the  input  is  always  greater  than  the  output. 
No  work  is  gained.  What  is  gained  is  the  ability  to  exert  a  large 
force  through  a  small  distance  by  means  of  a  convenient  small 
force  acting  through  a  larger  distance,  or  a  large  displacement 
with  a  small  force  by  means  of  a  smaller  displacement  with  a 
larger  force.  One  may  not  be  able  to  lift  400  pounds  vertically 
upward  a  distance  of  2  feet,  but  he  may  easily  accomplish  the 
same  result  by  means  of  an  inclined  plane  10  feet  in  length. 

Problems 

1.  A  barrel  weighing  240  pounds  is  lifted  a  vertical  distance  of  3  feet  by 
rolling  up  a  plank  12  feet  in  length  as  an  inclined  plane.     The  force  is 
applied  parallel  to  the  plank  by  means  of  a  frictionless  bearing  at  the  axis. 
If  the  rolling  friction  is  negligible,  what  is  the  work  done?     What  is  the 
force,  and  what  is  the  mechanical  advantage? 

2.  The  barrel  of  Problem  1  is  rolled  up  the  plane  by  means  of  a  rope  which 
is  wound  around  it  several  times  and  then  fastened.     How  far  will  the  end 
of  the  rope  move  while  the  barrel  rolls  12  feet,  if  the  pull  is  applied  parallel 
to  the  plane?    What  is  the  pull?     What  is  the  mechanical  advantage? 
What  is  the  minimum  value  of  the  coefficient  of  friction  between  the  barrel 
and  the  plank  in  order  that  it  may  be  rolled  in  this  way?     (The  diameter 
of  the  rope  is  negligible.) 

3.  A  body  is  pulled  up  a  30-degree  inclined  plane  by  means  of  a  force 
parallel  to  the  plane.     The  body  is  on  rollers  which  have  a  combined  fric- 
tion equal  to  a  coefficient  of  sliding  friction  of  0.02.     Find  the  efficiency. 

The  useful  work  is  W  s.  sin  30°  The  normal  pressure  is  W  cos  30°.  The 
lost  work  of  friction  is  0.02  W  s  cos  30° 

Useful  work  =  0.500  Ws; 
lost  work  =  0.0173  Ws; 

0.500 

efficiency  =  r~rr^  =  96.7  per  cent. 
0.517 

Frequently,  the  body  is  lifted  in  a  carriage  which  takes  up  part 
of  the  energy.  If  the  energy  of  the  carriage  is  not  returned  in  the 
form  of  useful  work,  it  represents  a  loss  which  must  be  taken  into 
account  in  calculating  the  commercial  efficiency  of  the  machine. 

Problems 

4.  A  mass  of  100  pounds  is  lifted  up  a  30-degree  plane  on  a  carriage 
which  weighs  20  pounds.     The  pull  is  by  means  of  a  rope  parallel  to  the 

14 


210 


MECHANICS 


[ART.    125 


plane.  The  coefficient  of  friction  is  0.03.  Find  the  commercial  efficiency 
of  the  inclined  plane  and  carriage  if  the  carriage,  when  running  back,  helps 
lift  another  one. 

Ans.     93.2  per  cent,  neglecting  the  loss  in  transmission  to  the  second 

carriage. 
6.  Solve  Problem  4  if  the  carriage  is  let  down  by  a  brake. 

Ans.     79.2  per  cent. 

6.  If  the  efficiency  is  95  per  cent,  what  load  can  be  lifted  up  a  16-degree 
inclined  plane  by  a  pull  of  100  pounds  parallel  to  the  plane/ 

The  mechanical  advantage  is  the  cosecant  of  16  degrees.     If  the  efficiency 
were  100  per  cent, 

W  =  100  X  3.628  =  362.8  Ib. 
Allowing  for  the  losses, 

W  =  362.8  X  0.95  =  344.7  Ib. ' 

7.  If  the  efficiency  is  94  per  cent,  what  force  parallel  to  a  20-degree  in- 
clined plane  will  lift  600  pounds  up  the  plane? 

125.  Wheel  and  Axle.— When  a  wheel  of  radius  r  (Fig.  189) 
rotates  through  an  angle  of  6  radians,  the  linear  displacement  is 
rd.     If  the  wheel  of  radius  r  is  attached  to 
an  axle  of  radius  a,  the  equation  of  work  is 


WaS  =  PrB; 
W_  r 
~P~  a 


(1) 
(2) 


Fia/189. 


The  mechanical  advantage  of  the  wheel  and 
axle  is  the  ratio  of  the  radius  (diameter)  of 
the  wheel  to  the  radius  (diameter)  of  the 
axle. 

Equation  (2)  might  be  derived  by  mo- 
ments about  the  axis.  A  wheel  and  axle  may 
be  regarded  as  a  continuous  lever. 


Problems 

1.  The  diameter  of  a  wheel  is  2  feet,  and  the  diameter  of  the  axle  is  6 
inches.     The  rope  on  the  wheel  is  %  inch  in  diameter  and  the  rope  on  the 
axle  is  1  inch  in  diameter.     Find  the  mechanical  advantage. 

Ans.     3.5. 

2.  A  wheel  40  inches  in  diameter  is  attached  to  an  axle  8  inches  in  diam- 
eter.    The  bearings  are  3  inches  in  diameter  and  the  coefficient  of  friction 
is  0.03.     The  wheel  and  axle  together  weigh  120  pounds.     Both  ropes  run 
vertically  downward.     Neglecting  the  diameter  of  the  ropes,  find  the  force 
required  to  lift  200  pounds,  and  find  the  efficiency. 

Ans.    40.8  Ib.;  98  per  cent. 


CHAP.   X1I1] 


WORK  AND  MACHINES 


211 


3.  Solve  Problem  2  for  a  load  of  100  pounds  and  for  a  load  of  600  pounds. 

4.  What  force  is  required  with  a  wheel  4  feet  in  diameter  and  an  axle 
10  inches  in  diameter  to  lift  a  load  of  1200  pounds,  if  the  rope  on  the  wheel 
is  0.5  inch  in  diameter;  the  rope  on  the  axle  is  1  inch  in  diameter;  and  the 
efficiency  is  94  per  cent? 

6.  A  body  is  pulled  up  a  20-degree  inclined  plane  by  means  of  a  rope 
parallel  to  the  plane.  The  rope  runs  from  an  axle  12  inches  in  diameter 
which  is  turned  by  a  wheel  40  inches  in  diameter.  Neglecting  the  diameter 
of  the  ropes,  find  the  mechanical  advantage  of  the  combination.  If  the 
combined  efficiency  is  92  per  centj  what  force  on  the  wheel  will  lift  1200 
pounds  up  the  plane? 

126.  Pulleys. — Figure  190  shows  a  single  pulley  with  a  flexible 
cord.  When  the  effort  P  moves  downward  a  distance  s,  the 
load  W  moves  upward  an  equal  distance.  The  mechanical 
advantage  is  unity. 


w 


FIG.  190. 


FIG.   191. 


In  Fig.  191,  there  is  a  fixed  pulley  at  the  top  and  a  single 
movable  pulley  below.  If  the  movable  pulley  is  lifted  a  distance 
s,  the  rope  supporting  it  is  shortened  that  distance  on  each  side 
and  the  part  of  the  rope  which  passes  over  the  fixed  pulley  must 
travel  a  distance  of  2s. 

P2s  =  Ws; 


The  mechanical  advantage  of  a  single  movable  pulley  supported 
by  two  parts-  of  a  continuous  rope  is  2.  This  is  evident  from  the 
fact  that  the  tension  in  a  continuous  rope  which  runs  over  smooth 
pulleys  is  the  same  in  all  parts  of  its  length,  and  two  parts  of 


212 


MECHANICS 


[ART.   126 


the  rope  support  the  load  W,  while  only  one  part  supports  the 
force  P. 

Since  the  block  and  pulley  of  weight  TFi,  which  form  the  mov- 
able system,  must  be  lifted  along  with  the  load  Wz,  the  com- 


mercial efficiency  is 


W 


if  there  are  no  losses  due  to  friction 


w, 


w, 


of  the  pulleys  or  bending  of  the  ropes. 

In  Fig.   192,  the  movable 
block  carries  one  pulley  and 

?   .     I  &  I        the   fixed    block   carries  two 

pulleys.  The  rope  is  attached 
to  the  movable  block.  For 
clearness,  the  pulleys  in  the 
upper  block  are  made  of  un- 
equal size  and  placed  one 
above  the  other.  This,  how- 
ever, is  not  a  customary 
arrangement.  It  is  evident 
that  the  mechanical  advan- 
tage is  3  when  the  blocks  are 
so  far  apart  that  the  three 
ropes  which  support  the 
movable  block  are  practically 
parallel.  In  the  position 
shown  in  Fig.  192,  the  me- 
chanical advantage  is  less 
than  3. 

Figure  193  shows  a  system 

with    two    pulleys    in    the    movable   block.     The   mechanical 

advantage  is  4  when  the  blocks  are  so  far  apart  that  the  ropes 

are  practically  parallel.     The  pulleys  on  each  block  run  on  the 

same  axis  but  with  different  speeds. 

In  a  system  of  pulleys  with  one  continuous  rope,  the  mechanical 

advantage  is  equal  to  the  number  of  ropes  which  directly  support 

the  weight. 

Problems 

1.  Sketch  a  system  of  pulleys  arranged  as  in  Fig.  192  with  a  mechanical 
advantage  of  5. 

2.  In  a  system  similar  to  Fig.  193,  the  mechanical  efficiency  is  95%. 
The  movable  block  and  pulleys  weigh  20  pounds.     The  weight  TF2  is  400 


FIG.  192. 


FIG.  193. 


CHAP,  X11I]  WORK  AND  MACHINES  213 

pounds.     Find  the  force  required  to  lift  the  system,  and  find  the  commercial 
efficiency. 

3.  A  safe  weighing  2,400  pounds  is  pulled  up  an  inclined  plane  by  a  block 
and  tackle.     The  plane  is  24  feet  long  and  5  feet  high.     The  block  and 
tackle  has  a  mechanical    advantage  of  4  and  an  efficiency  of  95  per  cent. 
The  efficiency  of  the  plane  is  90  per  cent.     Find  the  force  and  find  the  re- 
quired length  of  rope 

4.  Sketch  a  system  of  pulleys  combined  with  a  wheel  and  axle  for  a 
combined  mechanical  advantage  of  12. 

127.  The  Screw.— In  lifting  a  body  by  means  of  a  screw,  the 
effort  travels  a  distance  2irr  in  one  revolution,  where  r  is  the 
distance  of  the  effort  from  the  axis.     At  the  same  time,  the  load 
moves  a  distance  equal  to  the  pitch  of  the  screw.     If  d  is  the 
pitch  (distance  between  threads), 

Wd  =  2-irPr; 
mechanical  advantage  =  —-=-• 

The  efficiency  of  a  screw  is  small.  It  must  be  less  than  50 
per  cent  or  the  screw  will  run  down  when  the  turning  moment  is 
released.  The  screw  is  a  continuous  inclined  plane  with  the 
force  applied  parallel  to  the  base.  If  the  friction  on  an  inclined 
plane  is  less  than  the  component  of  the  load  parallel  to  the  plane, 
the  load  will  slide  down.  In  order  to  pull  a  body  up  a  plane  where 
the  friction  is  sufficient  to  prevent  it  from  sliding  down,  the  force 
parallel  to  the  plane  must  be  twice  as  great  as  the  component  of 
the  weight.  The  force  parallel  to  the  base  must  be  still  greater. 

Problem 

A  screw  with  4  threads  to  the  inch  is  turned  by  a  lever  15  inches  long 
from  the  axis  of  the  screw  to  the  point  of  application  of  the  effort.  What  is 
the  mechanical  advantage?  If  the  efficiency  is  30  per  cent,  what  is  the 
force  required  to  lift  2,000  pounds?  .  Ans.  377,17.7  Ib. 

128.  Differential  Appliances. — Figure  194  shows  a  differentia 
screw.     The  part  which  passes  through  the  nut  A  is  made  with 
one  pitch,  while  the  part  which  passes  through  the  nut  B  is  made 
with  a  smaller  pitch.     If  the  screw  is  rotated  in  a  clockwise  direc- 
tion, (as  seen  by  an  observer  looking  toward  the  head  from  the 
right),  it  will  move  toward  the  left  through  the  nut  A.     If  the 
nut  B  is  held  so  that  it  does  not  rotate  with  the  screw,  it  will 


214 


MECHANICS 


[ART.    128 


move  toward  the  right  relative  to  the  screw,  but  will  move 
toward  the  left  relative  to  the  nut  A.  The  actual  motion  of  B 
relative  to  A  during  one  revolution  is  equal  to  the  difference  of 
pitch  of  the  two  parts  of  the  screw. 


>— -^ 


FIG.  194. 

Problems 

1.  The  pitch  of  the  right  end  of  the  screw  of  Fig.  194  is  three  threads 
to  the  inch  and  the  pitch  of  the  left  end  is  four  threads  to  the  inch.     The 
screw  is  turned  by  means  of  a  rod  which  is  20  inches  long  from  the  axis  of 
the  screw  to  the  point  of  application  of  the  force.     Find  the  mechanical 
advantage.  Ans.     1508. 

2.  What  would  be  the  mechanical  advantage  in  Problem  1  if  the  pitches 
were  8  threads  and  9  threads  to  the  inch,  respectively? 

Figure  195  shows  a  differential  axle.  As 
part  of  the  rope  around  the  axle  of  radius 
a2  is  wound  up,  the  part  around  the  axle 
of  radius  0,1  is  unwound.  In  turning 
through  an  angle  of  6  radians,  the  total 
vertical  rope  is  decreased  in  length  by  an 
amount  (a2  —  a\)Q  and  the  load  is  lifted 

-^-s — -  0.     The   effort    moves   through   a 

z 

distance  r6,  and  mechanical  advantage  = 
—     The   combination   is  equivalent 

Cf2  —   fli 

to  an  axle  of  radius  a2  —  ai  lifting  a  single 
movable  pulley  with  two  ropes  to  support 
the  weight.  With  this  arrangement,  the 
mechanical  advantage  of  a  small  axle  is 
secured  and  the  axle  is  sufficiently  large  to 
FIG.  195.  carry  the  required  load. 

Problems 

3.  In  Fig.  195,  the  wheel  is  20  inches  in  diameter.     The  larger  part  of 
the  axle  is  6  inches  in  diameter  and  the  smaller  part  is  5  inches  in  diameter. 
Find  the  mechanical  advantage. 


CHAP.   XIII]  WORK  AND  MACHINES  215 

4.  Find  the  mechanical  advantage  of  Problem  3  by  moments  about  the 
axis  of  the  axle. 

In  many  machines  it  is  not  easy  to  determine  the  mechanical 
advantage  from  inspection.  The  mechanical  advantage  may 
always  be  found  by  moving  the  machine  and  measuring  the 
relative  motion  of  the  effort  and  the  resistance.  In  a  differential 
pulley,  for  instance,  it  was  found  that  the  chain  by  which  the 
effort  was  applied  moved  50  inches,  while  the  hook  which  carried 
the  load  moved  1  inch.  Evidently  the  mechanical  advantage 
is  50. 

129.  The  Lever. — The  mechanical  advantage  of  the  lever,  is 
most  easily  calculated  by  moments.     If  the  forces  are  perpendicu- 
lar to  the  moment  arms,  as  in  Fig.  196,   I;  or  if  the  lever  is 
straight  so  that  the  points  of  application  of  all  the  forces  are  on 
the  same  straight  line,  and  the  forces  are  parallel,  the  moment 
equation  is 

Pa  =  Wb,  (1) 

.?-5'  .      & 

Equation  (2)  states  that  the 
mechanical  advantage  is  the 
ratio  of  the  effort  arm  to  the 
resistance  arm. 

If  the  lever  of  Fig.  196,  I, 
is  turned  through  a  small 
angle  d0,  the  work  done  by 
the  force  P  is  PadQ,  and  the 
negative  work  of  the  resis- 
tance W  is  WbdB.  Equation  FlG-  196. 
(1)  is  obtained  by  equating  these  two  expressions  for  work. 

In  Fig.  196,  II,  the  work  of  the  force  P  is  Pad6  cos  0,  and  the 
work  of  the  force  W  is  W bd9  cos  6,  in  which  6  is  the  angle  between 
the  lever  and  a  line  perpendicular  to  the  forces. 

The  efficiency  of  a  lever  as  a  machine  may  be  nearly  100 
per  cent,  when  the  fulcrum  is  a  carefully  made  knife  edge  as  in 
a  chemical  balance. 

130.  Virtual  Work.— The  preceding  articles  of  Chapter  XIII 
have  dealt  with  bodies  moving  with  uniform  speed.     The  same 
conditions  apply  to  a  body  at  rest.     If  a  body  in  equilibrium 
be  considered  as  suffering  a  small  displacement,  the  total  work  of 


216 


MECHANICS 


[ART.   130 


all  the  forces  will  be  zero.  In  Fig.  197,  the  body  of  mass  W 
which  is  held  on  a  smooth  inclined  plane  by  a  force  P  parallel  to 
the  plane  may  be  regarded  as  moved  up  the  plane  a  distance  ds. 
The  body  is  raised  a  vertical  distance  dy,  so  that 


-  Wdy  +  Pds  =  0 


P  = 


Wdy 
ds 


W  sin  6. 


(1) 
(2) 


Figure  198  shows  a  bar  of  length  I,  mass  W,  with  its  center  of 
mass  at  a  distance  a  from  the  lower  end.     The  bar  rests  on  a 
smooth  horizontal  floor  and  leans   . 
against    a    smooth    vertical    wall. 
The  bar  is  held  from  slipping  by  a 
horizontal    force    at   the    bottom. 
This  horizontal  force  will  be  com- 
puted by  the  method  of  work. 


FIG.  197. 


FIG.  198. 


When  the  lower  end  of  the  bar  is  moved  a  distance  dx  toward 
the  right,  the  center  of  mass  is  moved  upward  a  distance  dy.  As 
the  reactions  at  the  wall  and  the  floor  are  normal  to  the  displace- 
ment, their  work  is  zero.  This  work  equation  is 


Pdx  =  Wdy 


(3) 


The  relation  between  dx  and  dy  is  found  by  differentiating  an 
equation  between  x  and  y.  This  relation  is  conveniently  found 
by  means  of  a  third  variable.  If  6  is  the  angle  between  the  bar 
and  the  horizontal,  and  y  is  the  distance  of  the  center  of  mass 
above  the  floor, 


y  =  a  sin  6, 
dy  =  a  cos  dB. 


(4) 
(5) 


CHAP.   XIII] 


WORK  AND  MACHINES 


217 


If  x  is  the  distance  of  the  lower  end  of  the  bar  from  the  vertical 
wall  and  is  positive  toward  the  right, 

x  =  —  I  cos  8,  (6) 

dx  =  I  sin  B  dO.  (7) 

Substitution  in  Equation  (3)  gives 

PI  sin  0  d6  =  Wa  cos  6  dO,  (8) 

W   _  a  cotan  6 
P  =  1 


(9) 


This  method  of  determining  the  conditions  of  equilibrium 
by  assuming  that  the  body  suffers  a  small  displacement  and 
equating  the  work  of  all  the  forces  is  called  the  method  of  virtual 
work. 

Problems 

1.  A  ladder  20  feet  long,  weighing  36  pounds,  with  its  center  of  gravity 
8  feet  from  the  lower  end,  rests  on  a  smooth  horizontal  floor  and  leans 
against  a  smooth  vertical  wall.     It  is  held  from  slipping  by  a  horizontal 
force  of  16  pounds  applied  2  feet  from  the  bottom.     Find  the  position  of 
equilibrium  by  the  method  of  virtual  work.     Check 

by  moments  and  resolutions. 

Ans.     45  degrees  with  the  horizontal. 

2.  The  ladder  of  Problem  1  rests  on  a  smooth 
horizontal   floor   and   leans   against  a  smooth  wall 
which  makes  an  angle  of  15  degrees  with  the  vertical, 
away  from  the  ladder.     It  is  held  from  slipping  by  a 
horizontal  push  of  8  pounds  at  the  bottom.     Find 
the  position  of  equilibrium. 

Ans.     Tan  6  =  1.5321;  6  =  56°  52'  with  the 
horizontal. 

3.  Find  the  mechanical  advantage  of  the  toggle 
joint,  Fig.  199,  in  terms  of  x  and  the  lengths  a  and  b. 


y  = 
dy  = 

W 

Va2  -  **  +  \ 
—  x  dx 

/W  -  xz, 
-  xdx 

Va2  -  x2       A 

/bz  -  x2 

V(o2  -  xz)(b2 

-  x2) 

P                       xy 

4.  In  Fig.  199,  x  =  2  feet;  a  =  b  =  8  feet.     Find 
the  mechanical  advantage. 


FIG.  199. 


Example 

Figure  200  shows  the  crank,  connecting  rod,  and  cross-head  of  an  engine. 
Find  the  relation  of  the  total  piston  pressure  P  to  the  component  of  the 


218 


MECHANICS 


[ART.    130 


force    perpendicular    to    the    crank.     Give    the    result    in    terms    of   the 
lengths  of  the  crank  and  the  connecting  rod  and  the  position  of  the  crank. 

x  =  a  cos  0  +  6  cos  <f>  =  a  cos  0  +  \/62  —  a2  sin20; 
-P  dx  =  R  a  de\ 


p  =  sin 


a  sin  0  cos  0 


sn 


cos  e  tan 


Problems 

6.  In  Fig.  200,  find  the  normal  reaction  N  at  the  cross-head  by  resolutions, 
Then  find  the  moment  about  the  axis  of  the  axle  at  0  by  a  moment  equation. 


FIG.  200. 

regarding  the  connecting  rod  and  crank  as  a  single  body.     Compare  the 
result  with  that  obtained  by  means  of  the  answer  of  the  Example. 

6.  A  bar  AB,  Fig.  201,  of  length  c  is  hinged  at  A  and  supported  by  a  rope 
at  B.  The  rope  runs  over  a  smooth  pulley  C  at  the  same  level  as  A,  at  a 
distance  6  therefrom,  and  carries  a  load  of  P  pounds  on  the  free  end.  The 
bar  weighs  W  pounds  and  its  center  of  mass  is  at  a  distance  a  from  the 
hinge.  Find  the  ratio  of  W  to  P  at  any  position. 


FIG.  201. 

Let  BC  =  s,  and  let  the  vertical  distance  of  the  center  of  mass  below  the 
hinge  be  represented  by  y,  which  is  taken  as  positive  downward. 
Wdy  =  P  ds; 

T7    =   be  sin  0 

p        a  cos  0V&2  +  c2  -  26c  cos  0' 

7.  In  Problem  6,  let  W  =  12  pounds,  P  =  10  pounds,  a  =  4  feet,  6  = 
8  feet,  and  c  =  6  feet.     Find  the  equation  for  the  angle  0. 

Ans.     0.96  cos3  0-2  cos2  0  +  1=0. 


CHAP.   XIII]  WORK  AND  MACHINES  219 

8.  Solve  Problem  7  for  cos  6  by  the  method  of  trial  and  error. 

cos  6  0.96  cos3  6  —  2  cos2  6  f(6) 

1  0.96  -2  -0.04 

00.  0  +1 

The  function  of  6  changes  sign  between  cos  0  =  0  and  cos  6  =  1.  There 
must  be  one  real  root  or  three  real  roots  between  these  values.  One  root 
is  evidently  much  closer  to  unity  than  to  zero.  Beginning  with  cos  0  =  0.9, 


0.9  0.69984 

0.95  0.82308 

0.96  0.84936 

0.97  0.87616 

0.965  0.862687 

0 . 966  0 . 865372 


.62  -  +0.07984 

.805  +0.01808 

.84320  +0.00616 

.88180  -0.00564 

.862450  +0.000237 

.866312  -0.000940 


An  interpolation  of  the  last  two  results  gives  cos  6  =  0.9652,  6  =  15°  10'. 

The  remaining  roots  may  be  found  by  a  similar  process.  It  is  better, 
however,  to  divide  the  cubic  equation  by  cos  0  —  0.9652.  The  quotient 
is  an  equation  of  the  second  degree,  which  may  be  solved  by  the  ordinary 
methods  for  quadratics.  One  of  the  roots  of  this  quadratic  is  a  solution  of 
the  problem  of  mechanics.  The  other  root  is  a  solution  of  the  mathematical 
equation,  but  is  geometrically  impossible.  Draw  the  space  diagram  for  the 
possible  result. 

9.  A  straight  bar  of  length  I  and  mass  W,  with  its  center  of  mass  at  a 
distance  a  from  the  lower  end,  rests  on  a  smooth  horizontal  floor  and  leans 
over  the  edge  of  a  smooth  wall.     The  vertical  height  from  the  floor  to  the 
point  of  contact  with  the  wall  is  h.     A  rope  attached  to  the  bottom  of  the 
bar  runs  horizontally  over  a  smooth  pulley  and  supports  a  mass  of  P  pounds. 
Find  the  equation  for  the  position  of  equilibrium  by  the  method  of  virtual 
work.  Ans.     Wa(cos  6  —  cos3  6)  =  Ph. 

10.  Solve  Problem  9  by  trial  and  error  for  the  following  cases: 


a 

h 

W 

P 

20' 

10' 

12' 

15# 

4# 

16' 

10' 

12' 

30# 

8# 

20' 

10' 

12' 

io# 

5# 

In  many  problems  which  may  be  solved  by  virtual  work,  there 
is  only  one  term  in  the  work  equation.  For  equilibrium,  this 
term  must  equal  zero.  If  the  force  in  this  term  is  the  weight  of 
some  body,  the  work  is  zero  when  the  point  of  application  of  this 
force  is  moving  in  a  horizontal  direction.  The  path  of  the  point 
of  application  is  a  curve  of  some  form.  If  the  curve  is  concave 
upward  at  the  point  at  which  the  tangent  is  horizontal,  the  equi- 
librium is  stable.  If  it  is  concave  downward,  the  equilibrium  is 
unstable. 


220 


MECHANICS 
Problems 


[ART.   131 


11.  A  rod  of  length  I,  Fig.  202,  with  its  center  of  mass  at  a  distance  a 
from  one  end,  slides  inside  a  smooth  hemispherical  bowl  of  radius  r.  Find 
the  position  of  equilibrium  if  I  is  less  than  2r. 

It  is  evident  that  the  path  of  the  center  of  mass  is  a  circle  with  its  center 
at  the  center  of  the  sphere.  Equilibrium  exists  when  the  center  of  mass  is 
at  its  lowest  point  directly  under  the  center  of  the  sphere.  In  this  position, 
find  the  angle  which  the  bar  makes  with  the  horizontal. 


FIG.  203. 


12.  Solve  Problem  11  when  I  =  6  inches,  a  =  2  inches,  and  r  =  4  inches. 

Ans.     Bar  makes  an  angle  of  20°  42'  with  the  horizontal. 

13.  A  bar  AB,  Fig.  203,  is  8  feet  long  and  has  its  center  of  mass  5  feet  from 
A.     The  end  B  rests  against  a  smooth  vertical  wall.     The  end  A  is  supported 
by  a  cord,  10  feet  in  length,  which  is  fastened  to  the  vertical  wall  directly 
over  the  end  B.     Find  the  position  of  equilibrium.     Solve  by  virtual  work 
and  check  by  the  direction  condition  of  equilibrium. 

Ans.     The  bar  makes  an  angle  of  53°  05'  with  the  vertical. 

131.  Character  of  Equilibrium. — It  is  often  easy  to  determine 
the  kind  of  equilibrium  by  means  of  the  methods  of  work.  In 
Problem  1 1  of  the  preceding  article,  the  path  of  the  center  of  mass 
is  a  circle  and  equilibrium  occurs  at  the  lowest  point.  Work 
must  be  done  on  the  bar  to  lift  its  center  of  mass  in  either  direc- 
tion up  the  curve.  If  the  center  of  mass  is  displaced  up  the 
curve,  the  force  of  gravity  will  do  work  on  it  in  bringing  it  back 
to  the  position  of  equilibrium.  The  equilibrium  is  stable. 

If  the  bar  of  Problem  11  were  inside  a  complete  sphere,  and  if 
the  ends  of  the  bar  passed  through  slots  in  the  surface  of  the 
sphere  and  were  bent  over  so  that  the  bar  could  slide  inside  the 
top,  another  point  of  equilibrium  could  be  found  at  the  top  of 
the  sphere.  The  path  of  the  center  of  mass  would  be  concave 


CHAP.   XIII] 


WORK  AND  MACHINES 


221 


downward  at  this  position.  If  the  bar  were  displaced  slightly 
from  this  position,  the  force  of  gravity  would  tend  to  displace 
it  still  farther,  and  it  would  not  return.  In  this  case,  the  equi- 
librium is  unstable.  In  many  cases,  when  a  body  is  displaced 
from  its  position  of  unstable  equilibrium,  it  will  move  to  the 
position  of  stable  equilibrium. 

A  position  at  which  the  total  work  of  a  small  displacement  is 
zero,  is  a  position  at  which  the  potential  energy  of  the  system  is  a 
maximum  or  a  minimum.  If  gravity  is  the  only  force  which 
does  work  on  a  body,  the  potential  energy  is  a  maximum  when 
the  path  is  concave  downwards,  and  a  minimum  when  the  path 
is  concave  upwards. 

In  Fig.  203,  the  location  of  the  center  of  mass  for  one  position 
on  each  side  of  the  position  of  equilibrium  is  represented  by  a 
small  circle.  It  is  evident  that  the  locus  of  the  center  of  mass 
is  concave  downward,  and  that  the  equilibrium  is  unstable. 

Figure  201  applies  to  Problem  6  of  Art.  130  for  one  position  of 
equilibrium.  At  the  other  position  of  equilibrium,  the  bar  is 
above  the  hinge  and  toward  the  left.  In  this  second  position, 
the  center  of  mass  of  the  bar  is  higher  than  that  of  Fig.  201. 
The  length  BC  is  also  greater;  hence  the  load  P.  is  higher.  Conse- 
quently the  potential  energy  in  the  second  position  is  greater 
than  in  Fig.  201.  Figure  201  shows  the  position  of  stable 
equilibrium.  At  the  other  position,  the  equilibrium  is  unstable. 

132.  The  Beam  Balance. — Figure  204  shows  the  fundamental 
parts  of  a  beam  balance  with  equal  arms.  In  an  accurately 


constructed  balance,  the  three  knife  edges  are  in  the  same  plane 
and  the  center  of  gravity  of  the  beam  is  located  a  little  below  the 
central  knife  edge.  When  equal  loads  are  applied  to  the  end 
knife  edges,  the  resultant  of  these  loads  falls  on  the  central 
knife  edge  and  does  no  work  when  the  beam  is  deflected.  When 


222  MECHANICS  [ART.   132 

a  small  additional  load  p  is  placed  on  the  right  knife  edge,  the 
beam  is  turned  through  an  angle  0  in  a  clockwise  direction.  If 
W  is  the  weight  of  the  beam,  c  is  the  distance  of  its  center  of 
gravity  below  the  central  knife  edge,  and  a  is  the  distance  be- 
tween knife  edges,  a  moment  equation  gives 

We  sin  6  =  pa  cos  0;  (1) 

*»•-&  (2) 

For  small  angles,  the  tangent  is  equal  to  the  arc  in  radians  and 
equation  (2)  becomes, 


pa 
~~  We' 


(3) 


Equations  (2)  and  (3)  give  the  sensibility  of  the  balance.  The 
sensibility  of  a  balance  is  the  angle  through  which  the  beam  is 
turned  by  unit  additional  load  on  one  knife  edge.  Delicate 
balances  are  provided  with  a  pointer  which  projects  downward 
and  moves  in  front  of  a  graduated  scale.  The  sensibility  of  the 
balance  is  usually  expressed  in  terms  of  the  scale  reading. 

Problems 

1.  A  balance  beam  weighs  50  grams  and  its  center  of  gravity  is  0.008 
inch  below  the  central  knife  edge.     The  knife  edges  are  4  inches  apart  and 
are  in  the  same  plane.     The  pointer  is  10  inches  long.     How  much  will 
the  end  of  the  pointer  move  when  1  milligram  is  placed  on  one  pan? 

Ans.     0.10  inch. 

2.  The  center  of  gravity  of  a  balance  beam  is  frequently  adjusted  by 
means  of  a  nut,  which  may  be  turned  up  or  down  on  a  screw  at  the  top  of  the 
beam.     (This  screw  is  not  shown  in  Fig.  204.)     In  Problem  1,  the  beam 
weighs  48  grams  and  the  nut  weighs  2  grams.     How  much  must  the  nut 
be  moved  to  double  the  sensibility? 

If  the  central  knife  edge  is  above  the  plane  of  the  end  knife 
edges,  the  resultant  of  equal  loads  on  the  pans  is  a  load  of  2P 
units  directly  below  the  central  knife  edge.  In  order  to  deflect 
the  beam,  work  must  be  done  on  this  load.  The  sensibility  of 
such  a  balance  is  diminished  as  the  load  is  increased.  If  d  is  the 
distance  of  the  central  knife  edge  above  the  plane  of  the  end 
knife  edges,  the  moment  equation  becomes 

pa  cos  0  =  We  sin  6  +  2Pd  sin  0',  (4) 


tan 


CHAP.  XIII]  WORK  AND  MACHINES  223 

When  the  central  knife  edge  is  above  the  plane  of  the  end  knife 
edges,  the  equilibrium  is  stable  for  all  loads. 

Problem 

3.  In  the  beam  of  Problem  1,  the  central  knife  edge  is  0.0004  inch  above 
the  plane  of  the  end  knife  edges.  When  the  load  on  each  pan  is  50  grams, 
find  the  deflection  of  the  pointer  for  1  milligram  additional  load  on  one  pan. 

Ans.    0.091  inch. 

When  the  central  knife  edge  is  below  the  plane  of  the  end  knife 
edges  a  distance  d,  the  moment  of  the  resultant  force  2P  changes 
sign  and  Equation  (5)  becomes 


The  sensibility  is  increased  with  increased  load.  If  the  load 
is  too  great,  the  second  term  of  the  denominator  becomes  larger 
than  the  first  term  and  the  equilibrium  is  unstable. 

When  a  balance  is  loaded,  the  beam  is  slightly  bent.  The 
central  knife  edge  may  be  a  little  .  below  the  plane  of  the  end 
knife  edges  when  the  total  load  is  small  and  may  come  above  their 
plane  when  the  total  load  is  larger.  In  a  beam  of  this  kind,  the 
sensibility  may  increase  with  a  small  increase  of  total  load  and 
decrease  with  a  larger  load. 

The  theory  above  applies  to  balances  with  equal  arms.  The 
same  principles  obtain  when  the  arms  are  not  equal.  The  only 
difference  is  that  the  sum  of  the  loads  on  the  two  pans  replaces 
the  load  2P  of  the  formulas. 

The  arms  of  a  balance,  which  are  supposed  to  be  equal,  may 
not  be  equal.  Let  a  be  the  length  of  one  arm  and  b  the  length  of 
the  other.  Let  P  be  the  load  on  the  arm  of  length  a  which  bal- 
ances an  unknown  load  W  on  the  other  arm. 

Pa  =  Wb.  (7) 

Now  put  the  load  W  on  the  arm  of  length  a  and  suppose  that  it  is 
balanced  by  a  load  Q  on  the  other  arm. 

Qb  =  Wa,  (8) 

W  =  VPQ.  (9) 


The  true  weight  is  the  geometric  mean  of  the  apparent  weights. 
The  arithmetic  mean  of  the  apparent  weights  is  sufficiently  exact 
for  all  ordinary  purposes. 


224  MECHANICS  [ART.   133 

If  W  is  eliminated  from  Equations  (7)  and  (8), 


Problems 

4.  An  unknown  load  on  the  left  pan  of  a  balance  is  balanced  by  100.012 
grams  on  the  right  pan.     When  the  unknown  load  is  placed  on  the  right 
pan,  it  is  balanced  by  99.976  grams  on  the  left  pan.     Find  the  true  weight 
of  the  body.     What  is  the  difference  between  the  geometric  mean  of  Equa- 
tion (9)  and  the  arithmetic  mean? 

5.  Find  the  ratio  of  the  arms  in  Problem  4. 

133.  Surface  of  Equilibrium.  —  It  is  sometimes  desirable  to  find 
a  surface  of  such  form  that  a  body  under  a  given  set  of  forces 
will  be  in  equilibrium  at  any  point  on  it.  The  reaction  of  the 
surface  at  every  point  must  be  opposite  the  direction  of  the 
resultant  of  the  applied  forces.  This  means  that  the  resultant 
of  the  applied  forces  must  everywhere  be  normal  to  the  surface. 
Such  a  surface  is  called  an  equipotential  surface  for  the  forces 
involved.  The  surface  of  a  still  lake  is  an  equipotential  surface 
for  the  force  of  gravity. 

Since  the  resultant  force  is  normal  to  an  equipotential  surface, 
no  work  is  done  in  moving  a  body  from  one  part  of  the  surface  to 
another.  If  Hx  is  the  component  in  the  direction  of  the  X  axis 
of  the  resultant  of  all  the  forces  at  the  surface  (except  the  reaction 
of  the  surface),  the  work  done  in  displacing  a  body  a  distance 
dx  is  Hxdx.  If  V  is  the  component  in  the  direction  of  the  Y  axis, 
and  Hz  is  the  component  in  the  direction  of  the  Z  axis,  the  total 
work  of  a  displacement  in  the  surface  is 

U  =  Hxdx  +  Vdy  +  Hzdz  =  0.  (1) 

Example 

The  centrifugal  force  of  a  rotating  body  varies  as  the  distance  from  the 
axis  of  rotation.  A  body  of  liquid  rotates  at  constant  speed  about  a  vertical 
axis.  Find  the  equation  of  the  intersection  of  the  surface  of  the  liquid  with 
the  XY  plane. 

The  horizontal  force  on  a  unit  of  mass  is  kx  and  the  vertical  force  is  —  1. 
Equation  (1)  becomes, 

kxdx  —  Idy  =  0, 
dy  =  kxdx. 
kx*   ,   n 

y  =  ~T  +  c- 

The  curve  is  a  parabola  with  the  axis  vertical. 


CHAP.  XIII]  WORK  AND  MACHINES  225 

Problems 

1.  A  mass  of  10  pounds  slides  on  a  smooth  wire  in  a  vertical  plane.     The 
mass  is  attached  to  a  cord  which  passes  over  a  pulley  in  the  same  vertical 
plane.     The  other  end  of  the  cord  is  attached  to  a  spring.     A  force  of  1 
pound  stretches  this  spring  1  inch.     When  there  is  no  elongation  of  the 
spring,  the  end  of  the  cord  just  reaches  the  pulley.     If  the  wire  passes 
through  a  point  12  inches  above  the  pulley,  find  its  equation  in  order  that  the 
mass  may  be  in  equilibrium  at  any  point  on  it. 

The  tension  in  the  cord  is  equal  to  —  r  where  r  is  the  distance  of  the  end  of 
the  cord  from  the  pulley.  The  horizontal  component  of  this  tension  is  the 
total  tension  multiplied  by  the  cosine  of  the  angle  which  the  cord  makes 

with  the  horizontal.     This  cosine  is  -• 

Hx  =  -x;     V  =  -y  -  10; 

xdx  +  (y  +  W)dy  =  0; 

z2  +  ?/2  +  20y  =  C. 

When  x  =  0,  y  =  12;  C  =  384; 

xz  +  (y  +  10)  2  =  222. 

The  wire  is  in  the  form  of  a  circle  of  22  inches  radius  with  its  center  10  inches 
below  the  pulley. 

2.  Solve  Problem  1  if  the  end  of  the  cord  is  4  inches  from  the  pulley 
when  the  tension  in  it  is  zero.     (The  pull  =  —  (r  —  4).) 


Ans.     x*  +  y*  +  2Qy  -  4\x*  +  y*  =  336. 

3.  A  body  slides  on  a  curved  wire  in  a  horizontal  plane.  It  is  fastened 
to  two  cords.  One  cord  passes  over  a  pulley  at  A  and  is  attached  to  a 
spring  which  requires  a  pull  of  k\  pounds  to  stretch  it  unit  distance.  The 
second  cord  passes  over  a  pulley  at  B  and  is  attached  to  a  spring  which 
requires  a  pull  of  fc2  pounds  to  stretch  it  unit  distance.  The  points  A  and  B 
are  at  a  distance  c  apart  in  the  direction  of  the  X  axis.  Find  the  equation 
of  the  wire  if  the  body  is  in  equilibrium  at  any  point  on  it. 

Ans.     (fcj  +  /c2)(z2  +  i/2)  -  2k,cx  =  C 

in  which  the  origin  of  coordinates  is  located  at  A,  and  the  point  B  lies  on 
the  X  axis  to  the  right  of  A. 

134.  Summary.  —  The  amount  of  work  done  by  a  force  is  the 
product  of  the  magnitude  of  the  force  multiplied  by  the  compo- 
nent of  the  displacement  of  its  point  of  application  in  the  direc- 
tion of  the  force 

U  =  P  X  s  cos  a.  =  P  cosa  X  s.  Formula  III 

The  second  form  of  the  formula  states  that  work  is  the  product 
of  the  displacement  multiplied  by  the  component  of  the  force 
in  the  direction  of  the  displacement. 
When  the  direction  or  magnitude  of  the  force  varies,  or  when 

15 


226  MECHANICS  [ART.   134 

the  direction  of  the  displacement  varies,  the  increment  of  work  is 
given  by 

dU  =  P  cos  a  ds. 

A  machine  transmits  energy.  The  ratio  of  the  work  done  by 
a  machine  to  the  work  done  on  it  is  called  the  efficiency  of  the 
machine.  The  ratio  of  the  force  exerted  by  a  machine  to  the 
force  applied  to  it  is  called  the  mechanical  advantage  of  the  machine. 
The  efficiency  of  a  machine  is  never  quite  equal  to  unity.  The 
mechanical  advantage  may  have  any  value. 

Problems  of  equilibrium  may  be  solved  by  means  of  work  equa- 
tions. In  the  method  of  virtual  work,  the  body  or  system  is 
assumed  to  suffer  a  small  displacement  and  the  work  of  this 
displacement  is  equated  to  zero. 

Equilibrium  may  be  stable,  unstable,  or  neutral.  The  poten- 
tial energy  of  a  body  in  stable  equilibrium  is  a  minimum.  Work 
must  be  done  on  the  body  to  move  it  from  the  position  of  equilib- 
rium. The  potential  energy  of  a  body  in  unstable  equilibrium 
is  a  maximum.  If  the  body  is  slightly  displaced  from  the  position 
of  equilibrium,  work  must  be  done  on  it  to  move  it  back  to 
that  position.  When  a  body  is  in  neutral  equilibrium,  no  work 
is  required  to  move  it  in  either  direction. 

When  the  force  is  applied  parallel  to  an  inclined  plane,  the 
mechanical  advantage  is  the  ratio  of  the  length  of  the  plane 
to  its  height.  When  the  force  is  applied  parallel  to  the  base  of 
an  inclined  plane,  the  mechanical  advantage  is  the  ratio  of  the 
base  to  the  height. 

The  mechanical  advantage  of  a  wheel  and  axle  is  the  ratio  of 
the  diameter  of  the  wheel  to  the  diameter  of  the  axle. 

When  a  continuous  rope  is  used,  the  mechanical  advantage 
of  a  system  of  pulleys  is  equal  to  the  number  of  parts  of  the 
rope  which  support  the  weight. 

The  mechanical  advantage  of  a  screw  is  the  ratio  of  the  cir- 
cumference of  the  circle  in  which  the  effort  moves  to  the  pitch 
of  the  screw.  The  efficiency  of  a  screw  is  less  than  50  per  cent. 

The  mechanical  advantage  of  a  lever  is  the  ratio  of  the  length 
of  the  effort  arm  to  the  length  of  the  resistance  arm.  The 
efficiency  of  a  lever  may  be  very  high. 

The  sensibility  of  a  balance  is  proportional  to  the  length  of  the 
arms  and  inversely  proportional  to  the  mass  of  the  beam  and  the 
distance  of  its  center  of  gravity  below  the  central  knife  edge. 


CHAP.  X1I1]  WORK  AND  MACHINES  227 

If  the  central  knife  edge  is  above  the  plane  of  the  end  knife  edges, 
the  sensibility  is  decreased  with  increased  load.  If  the  central 
knife  edge  is  below  the  plane  of  the  end  knife  edges,  the  sensi- 
bility is  increased  with  increased  load.  If  there  is  much  differ- 
ence in  this  direction,  the  equilibrium  may  be  unstable  with  a 
large  load. 

If  the  arms  of  a  balance  are  unequal,  the  load  must  be  weighed 
first  on  one  pan  and  then  on  the  other.  The  true  weight  is  the 
geometric  mean  of  the  two  weighings.  The  arithmetric  mean 
is  sufficiently  exact  in  a  fairly  good  balance. 

On  an  equipotential  surface 

Hxdx  +  Vdy  +  Hxdz  =  0. 


CHAPTER  XIV 


MOMENT  OF  INERTIA  OF  SOLIDS 

135.  Definition. — The  moment  of  inertia  of  a  solid  with  respect 
to  an  axis  may  be  defined  mathematically  as  the  sum  of  the  prod- 
ucts obtained  by  multiplying  each  element  of  mass  by  the  square 
of  its  distance  from  the  axis.     Expressed  algebraically, 

/  =  2wr2,  (1) 

in  which  7  is  the  moment  of  inertia,  m  is  an  element  of  mass,  and 
r  is  the  distance  of  the  element  from  the  axis.  The  element 
must  be  of  such  form  that  all  parts  are  at  the  same  distance  from 
the  axis.  If  the  elements  are  infinitesimal,  Equation  (1)  is 
written 

7  =  fr*dm,  Formula    XI 

in  which  dm  is  the  element  of  mass. 

Figure  205  shows  the  form  of  the  element  required  to  find  the 

moment  of  inertia  with  respect  to 
the  Z  axis  OZ.  The  element  BB' 
is  parallel  to  the  axis  OZ.  If  its 
transverse  dimensions  are  infini- 
tesimal, all  parts  of  the  element 
are  at  the  same  distance  from  OZ, 
within  infinitesimal  limits.  The 
element  may  also  be  infinitesimal 
parallel  to  the  axis  or  it  may  ex- 
—  X  tend  entirely  through  the  body  in 
this  direction  as  shown  in  Fig. 
205.  The  cross-section  of  the  ele- 
ment in  the  plane  perpendicular  to  the  axis  may  be  rectangular 
as  shown,  or  may  have  the  form  necessary  for  integration  with 
polar  coordinates.  The  element  may  be  a  hollow  cylinder  of 
radius  r  and  thickness  dr,  which  extends  through  the  body 
parallel  to  the  axis. 

136.  Moment  of  Inertia  by  Integration. — Figure  206  shows  one 
end  of  a  cylinder  of  radius  a.     It  is  desired  to  find  the  moment  of 

228 


FIG.  205. 


CHAP.  XIV]     MOMENT  OF  INERTIA  OF  SOLIDS 


229 


inertia  of  this  cylinder  with  respect  to  its  axis,  which  is  perpen- 
dicular to  the  plane  of  the  paper. 

Polar  coordinates  are  best  suited  for  this  problem.  The 
element  of  volume  parallel  to  the  axis  of  the  cylinder  has  the 
cross-section  dA  =  rdddr.  If  I  is  the  length  of  the  cylinder, 
the  element  of  volume  is 


dV  =  Irdedr. 
If  p  is  the  density,  or  mass  per  unit  volume, 

dm  =  plrdddr. 
I  =  frzdm  =  plffr*dddr. 


(1) 

(2) 
(3) 


role 


FIG.  206. 


FIG.  207. 


Integrating  first  with  respect  to  r  and  putting  in  the  limits  r  =  0 
and  r  =  a, 

I  =  P-\~  dO.  (4) 

Integrating  next  with  respect  to  B  and  putting  in  the  limits 
6  =  0  and  6  =  27r, 

7  -  *?•  Mo*  T'      .          (5) 

The  mass  of  the  cylinder  is 

m  =  plird2, 
which  substituted  in  Equation  (5)  gives 

I  =  ^f-  Formula  XII 

Figure  207  represents  a  rectangular  parallelepiped  of  length  I, 
breadth  6,  and  depth  d.  It  is  desired  to  find  its  moment  of 
inertia  with  respect  to  the  axis  of  symmetry  parallel  to  its  length. 

The  line  00'  is  the  axis.     The  element  of  volume  is  dV  =  Idxdy. 


230  MECHANICS  [ART.   136 

The  element  of  mass  is  dm  =  pldxdy.     With  0-0'  as  the  axis 
of  coordinates, 

/  =  Jr2dm  =  plff(x2  +  y^dxdy  (6) 

The  limits  for  x  are  -^  and  +2-     The  limits  for  y  are  —^  and 


/  = 


Problems 

1.  Find  the  moment  of  inertia  of  the  rectangular  parallelepiped  of  Fig. 
207  with  respect  to  the  axis  CC". 

2         2 
Ans.  I  = 


3 

2.  Figure  208  shows  a  triangular  prism  of  length  I.  The  cross-section 
is  a  right-angled  triangle.  Find  the  moment  of  inertia  with  respect  to  the 
edge  00'. 

AnS.  1      =    f»n/u/  -  fy  —     iiv  ~ 

3.  Find  the  moment  of  inertia  of  a  right  circular 
cone  of  height  h  and  diameter  2a  with  respect  to  its 
axis. 

Build  up  the  cone  of  a  series  of  flat  disks  parallel  to 
the  base  as  shown  in  Fig.  209.     The  radius  of  each  disk 
is  r  and  its  thickness  is  dy.     The  distance  y  may  be 
°  L___£__J     measured  from  the  base,  as  in  Fig.  209  or  from  the 
F      208  vertex.     The  expression  for  the  moment  of  inertia  of 

the  disk  is  written  by  means  of  Formula  XII.  The 
radius  r  is  expressed  in  terms  of  y  and  the  constants;  the  moment  of  inertia 
is  found  by  a  single  integration. 

T       pirha1       3wa2 

"To-     "To" 

4.  Solve  Problem  3  by  building  up  the  cone  of  hollow  cylinders  of  radius 
r  and  thickness  dr,  coaxial  with  the  cone.  (Fig.  210.) 

The  solutions  of  Problems  3  and  4  are  instances  of  the  integration  of  an 
element  of  volume  rdddrdy.  Integrating  6  first  gives  a  hollow  ring.  Inte- 
grating y  next  builds  the  rings  into  a  hollow  cylinder.  Integrating  r  last 
expands  these  cylinders  to  form  the  cone.  For  Fig.  209,  r  may  be  integrated 
first  and  6  next,  or  6  may  be  integrated  first  and  r  next.  In  either  case,  the 
result  is  the  thin  disk.  The  final  integration  with  respect  to  y  builds  up 
the  cone  of  a  series  of  such  disks. 

6.  Find  the  moment  of  inertia  of  a  sphere  with  respect  to  a  diameter, 

using  disks  as  elements.  Ans.     /  =  -^~ 

10 


CHAP.  XIV]     MOMENT  OF  INERTIA  OF  SOLIDS 


231 


6.  Solve  Problem  5  by  a  single  integration,  using  hollow  cylinders  as 
elements  of  volume. 

7.  By  means  of  Equation  (5)  Problem  5,  find  the  moment  of  inertia  of  a 
hollow  cylinder  of  inside  radius  b  and  outside  radius  a. 


FIG.  209. 


FIG.  210. 


8.  A  right  circular  cone,  10  inches  in  diameter,  and  12  inches  high,  has 
a  coaxial  cylindrical  hole,  4  inches  in  diameter,  cut  through  it.  By  means 
of  the  results  of  the  preceding  examples  and  problems,  find  the  moment  of 
inertia  of  the  remainder.  Ans.  I  =  615.6  trp  =  1934p. 


FIG.  211. 


FIG.  212. 


9.  Find  the  moment  of  inertia  of  a  homogeneous  solid  cylinder  of  length 
I  and  radius  a  with  respect  to  a  line  in  its  surface  parallel  to  its  axis.     The 
end  of  this  cylinder  is  shown  in  Fig.  211. 

10.  Find  the  moment  of  inertia  of  a  homogeneous  solid  cylinder  with 
respect  to  an  axis  parallel  to  the  axis  of  the  cylinder  at  a  distance  d  therefrom. 

Figure  212  shows  the  end  of  the  cylinder.     The  axis  of  inertia  is  perpen- 


232  MECHANICS  [ART.   137 

dicular  to  the  plane  of  the  paper.  For  simplicity,  the  origin  of  coordinates 
is  taken  at  the  center  of  the  circle.  The  distance  r  of  Formula  XI  is  r'  of 
the  figure.  This  distance  must  be  expressed  in  terms  of  r,  0,  and  d. 

Ans.     I  =  ml—  + 


y  11.  In  Problem  10,  find  the  moment  of  inertia  of  the  first  quadrant  alone. 
12.  Find  the  moment  of  inertia  with  respect  to  the  Z  axis  of  the  tetra- 

X  77  2, 

hedron  bounded  by  the  three  coordinate  planes,  and  the  plane  -  +  r  +  -  =1. 

Ans.     I 


137.  Radius  of  Gyration.  —  The  radius  of  gyration  of  a  body  is 
the  distance  from  the  axis  at  which  all  the  mass  of  the  body 
must  be  placed  in  order  that  the  moment  of  inertia  may  be  the 
same  as  that  of  the  actual  body.  It  is  defined  mathematically 
by  the  equation 

k2  =  -,  Formula  XIII 

ra 

in  which  k  is  the  radius  of  gyration. 

The  moment  of  inertia  of  a  homogeneous  solid  cylinder  with 

77?  Cl  CL^ 

respect  to  its  axis  is  ~Q~;  anc*  ^  =  "9"'  ^  a  h°m°geneous  solid 
cylinder  of  given  mass  is  replaced  by  a  thin  hollow  cylindrical 
shell  of  the  same  mass,  the  moment  of  inertia  of  the  shell  will 
equal  that  of  the  cylinder,  provided  the  radius  of  the  shell  is 
0.7071  of  the  radius  of  the  cylinder. 

Problems 

1.  Find  the  radius  of  gyration  of  a  homogeneous  solid  sphere  20  inches  in 
diameter.  Ans.     k  =  \/40  =  6.32  in. 

2.  Find  the  radius  of  gyration  of  a  homogeneous  hollow  cylinder,  of  which 
the  outside  diameter  is  16  inches  and  the  inside  diameter  is  10  inches. 

Ans.     k  =  6.67  in. 

3.  Find  the  radius  of  gyration  of  a  homogeneous  hollow  cylinder,  of 
which  the  outside  diameter  is  16  inches  and  the  inside  diameter  is  15  inches. 

Ans.     k  =  7.75  in. 

4.  Derive  the  expression  for  the  square  of  the  radius  of  gyration  of  a 
hollow  cylinder,  of  which  the  outside  radius  is  a  and  the  inside  radius  is  b. 

Am.  *..*+£ 

6.  Derive  the  expression  for  the  square  of  the  radius  of  gyration  of  a 
homogeneous  solid  cone  of  radius  a  with  respect  to  the  axis  of  the  cone. 

7,       3a2 
Ans.     fc2=-- 


CHAP.  XIV]     MOMENT  OF  INERTIA  OF  SOLIDS 


233 


138.  Transfer  of  Axis. — The  calculation  of  the  moment  of 
inertia  about  any  axis  is  frequently  a  laborious  process.  The 
work  may  often  be  simplified  by  means  of  a  transfer  of  the  axis. 
If  the  moment  of  inertia  of  a  body  is  known  with  respect  to  some 
axis,  its  moment  of  inertia  with  respect  to  any  other  axis  parallel 
to  that  axis  may  be  found  by  simple  algebraic  methods. 

In  Fig.  213,  the  moment  of  inertia  with  respect  to  the  line 
00'  is  desired.  Let  BB'  represent  any  element  of  the  body 
parallel  to  00'.  The  distance  from  00'  to  BB'  is  r. 

I  =  fr2dm.  (1) 

Suppose  that  the  moment  of  inertia  of  the  body  is  known  with 
respect  to  the  axis  CC',  and  suppose  that  CCf  is  parallel  to  00' 


FIG.  213. 

at  a  distance  d.     The  horizontal  component  of  d  is  a  and  the 
vertical  component  is  6. 

&  =  a2  +  b2.  (2) 

The  coordinates  of  the  element  BB'  with  respect  to  CC'  as  the 
Z  axis  are  x  and  y. 

r*  =  (a  +  x)2  +  (b  +  y)2.  (3) 

/  =  /(02  +  2ax  +  x2  +  &2  +  2by  +  y2)dm',  (4) 

/  =  /(a2  +  b*)dm  +  f(x2  +  y2)dm  +  2afxdm  + 

2bfydm.     (5) 

The  first  term  of  the  second  member  of  Equation  (5)  is 
(a2  +  b2)fdm  =  md2. 


234  MECHANICS  [ART.  138 

The  second  term  is  the  moment  of  inertia  with  respect  to  CC'. 
This  moment  of  inertia  may  be  represented  by  70. 

f(xi  +  y2)dm  =  70. 

The  third  term  is  lafxdm.  The  expression  xdm  is  the  moment 
of  the  element  with  respect  to  the  vertical  plane  through  CC'. 
The  expression  fxdm  is  the  moment  of  the  entire  volume  with 
respect  to  this  plane.  If  the  vertical  plane  through  CC'  passes 
through  the  center  of  gravity  of  the  body,  the  moment  is  zero. 
Similarly,  the  last  term  is  zero  if  the  horizontal  plane  through 
CC'  passes  through  the  center  of  gravity  of  the  body.  If  the 
line  CC'  passes  through  the  center  of  gravity  of  the  body,  the  last 
two  terms  of  Equation  (5)  vanish  and 

/  =  /0  +  md\  Formula  XIV 

in  which  70  is  the  moment  of  inertia  with  respect  to  an  axis 
through  the  center  of  gravity,  /  is  the  moment  of  inertia  with 
respect  to  a  parallel  axis  at  a  distance  d  from  the  center  of  gravity, 
and  m  is  the  mass  of  the  body. 

Formula  XIV  affords  a  convenient  method  of  finding  the 
moment  of  inertia  of  a  body  with  respect  to  any  axis,  if  the 
moment  of  inertia  is  known  with  respect  to  a  parallel  axis  through 
the  center  of  gravity.  If  the  moment  of  inertia  is  known  for  some 
axis  which  does  not  pass  through  the  center  of  gravity,  and  if  it 
is  desired  for  some  other  axis  which  does  not  pass  through  the 
center  of  gravity,  Formula  XIV  may  be  used  to  transfer  from  the 
first  axis  to  the  center  of  gravity  and  then  to  transfer  from 
the  center  of  gravity  to  the  second  axis. 

If  k0  is  the  radius  of  gyration  with  respect  to  an  axis  through 
the  center  of  gravity, 

70  =  mkl. 


If  k  is  the  radius  of  gyration  with  respect  to  the  axis  00', 


md2       7  ,        ,9  ^          ,    VT, 

kz  =  —  =  -  -  =  kl  +  d2.  Formula  XV 

m  m 

Problems 

1.  Solve  Problem  9  of  Art.  136  by  means  of  Formula  XIV. 

2.  Solve  Problem  10  of  Art.  136  by  means  of  Formula  XIV. 

3.  A  solid  sphere  is  12  inches  in  diameter.     Find  its  radius  of  gyration 
with  respect  to  an  axis  10  inches  from  its  center.  Ans.     k  =  10.70  in. 

4.  Find  the  radius  of  gyration  of  a  hollow  cylinder,  8  inches  outside 


CHAP.  XIV]     MOMENT  OF  INERTIA  OF  SOLIDS 


235 


diameter  and  5  inches  inside  diameter,  with  respect  to  an  axis  10  inches  from 
the  axis  of  the  cylinder. 

6.  A  circular  disk,  20  inches  in  diameter,  has  4  holes  drilled  through  it. 
The  diameter  of  each  hole  is  5  inches  and  the  center  is  4  inches  from  the 
center  of  the  disk.  Find  the  radius  of  gyration.  Ans.  k  .=  7.76  inches. 

6.  Knowing  the  moment  of  inertia  of  a  sphere  and  the  location  of  the 
center  of  gravity  of  a  hemisphere,  find  the  moment  of  inertia  of  a  hemisphere 
with  respect  to  a  tangent  in  a  plane  parallel  to  the  plane  which  bounds  the 
hemisphere.  Solve  by  means  of  Formula  XIV.  Solve  also  by  means  of 
Equation  (5). 

139.  Moment  of  Inertia  of  a 
Thin  Plate.  —  The  moment  of 
inertia  of  a  retangular  parallelo- 
piped  of  length  I,  breadth  6,  and 
thickness  d,  with  respect  to  one 
edge  parallel  to  its  length  (Fig. 
214)  is 

+  d 


T         i^ 
1  =  plod 


m 


\\IHlii, 

't    Y        l' 

\ 

\ 

c 

FIG.  214. 

±d2 

(See  Problem  1  of  Art.  136).  If  the  thickness  is  small  compared 
with  the  breadth,  its  square  may  be  relatively  negligible  and 
may,  therefore,  be  neglected.  For  instance,  if  b  =  100  inches 
and  d  =  1  inch,  the  error  made  by  dropping  d2  is  only  one  part 
in  ten  thousand.  The  moment  of  inertia  of  a  long,  slender  rod 
with  respect  to  an  axis  through  one  end  perpendicular  to  its 

I2 
length  is  m  ~->  and  the  moment  of  inertia  with  respect  to  an  axis 

I2 
through  the  middle  perpendicular  to  its  length  is  m      *     These 


car-* 


-dm 


formulas  are  derived  from 
the  results  for  a  parallele- 
piped by  neglecting  the 
thickness  and  interchang- 
ing I  and  b. 

Problems 

1.  Find  the  moment  of  in- 
ertia of  a  thin  circular  disk  of 
thickness  dz  with  respect  to  a 
diameter. 

If   the   horizontal   diameter, 
00'  Fig.  215,  is  taken  as  the 
axis  of  reference,  the  element  of  volume  parallel  to  this  diameter  is  2xdydz. 


Front 

FIG.  215. 


236 


MECHANICS 


[ART.    140 


The  thickness  dz  is  so  small  (many  times  less  than  shown  in  the  side  view 
of  Fig.  215)  that  its  square  may  be  neglected  in  calculating  r. 

r  =  y, 

I  =  pdzS2xy2dy, 

which  is  best  solved  by  substituting  x  =  a  cos  0,  y  =  a  sin  0. 

/sin2  20 
— 2 — d0' 


pa*dzf- 


,    Tr 
pdz  —     = 


—  cos  40 


d6  =  pa*dz 


[0      sin  401  2 
J--H-J  v 


This  moment  of  inertia  is  one  half  that  of  the  disk  with  respect  to  its  axis. 

2.  Find  the  moment  of  inertia  of  a  rectangular  plate  of  breadth  6,  length 
I,  and  thickness  dz,  with  respect  to  an  axis  through  its  center  parallel  to  its 

,         ,  7    b3       mb2 
length.  Ans.     I  =  pldz  —  =  -^  • 

3.  Find  the  moment  of  inertia  of  a  triangular  plate  of  height  h,  base  6, 
and  thickness  dz,  with  respect  to  an  axis  through  its  center  of  gravity  par- 
allel to  its  base. 

Find  the  moment  of  inertia  first  with  respect  to  an  axis  through  the  vertex 
parallel  to  the  base.  Transfer  to  a  parallel  axis  through  the  center  of  gravity 
by  means  of  Formula  XIV. 

Ans.     I 


36 


mfe2 
18  ' 


140.  Plate  Elements. — It  is  frequently  convenient  to  build  up  a 
solid  of  a  series  of  thin  plates.  The  moment  of  inertia  of  each 
plate  is  found  for  an  axis  through  its  center  of  gravity  and  then 
transferred  to  the  required  axis.  Finally,  the  moment  of  inertia 
of  the  entire  solid  is  found  by  integration. 

Example 

Find  the  moment  of  inertia  of  a 
homogeneous  solid  cylinder  of 
length  I,  and  radius  a  with  respect 
to  a  diameter  at  one  end.  (Fig. 
216.) 

From  Problem  1  of  Art.  138,  the 
moment  of  inertia  of  a  disk  of 
radius  a  and  thickness  dx  with 
respect  to  the  diameter  CC"  is 

ir-r-  dx.     The  moment  of  inertia  with  respect  to  the  parallel  axis  00'  at  a 
distance  x  from  CC'  is 


7  =  70  +  ma;2  =  pira*x*dx  +  pir-j-dx; 


(1) 


; 

CHAP.  XIV]     MOMENT  OF  INERTIA  OF  SOLIDS  237 


(3) 

Problems 

1.  Find  the  moment  of  inertia  of  a  homogeneous  solid  cylinder  with  respect 
to  a  diameter  at  the  middle.  Ans.     I  =  m  l^~  +  —  ]  * 

2.  Find  the  moment  of  inertia  of  a  circular  rod  with  respect  to  an  axis 
through  one  end  perpendicular  to  its  length,  if  the  diameter  is  so  small  that 
its  square  is  negligible  compared  with  the  square  of  the  length. 

,       mlz 
Ans.     I--—- 

3.  Find  the  moment  of  inertia  of  a  homogeneous  right  cone  of  altitude  h 
and  diameter  2a  with  respect  to  an  axis  through  the  vertex  parallel  to  the 

base.  Ans.     I  =  *  + 


4.  Solve  Problem  12  of  Art.  136  by  means  of  plate  elements. 
6.  Find  the  square  of  the  radius  of  gyration  of  a  square  right  pyramid 
with  respect  to  an  axis  through  the  vertex  parallel  to  one  edge  of  the  base. 

Ans.     jfc*  =  | 

For  any  right  pyramid  or  cone,  the  square  of  the  radius  of  gyration  with 
respect  to  an  axis  through  the  vertex  parallel  to  the  base  is  given  by  the 
equation, 

fc2  =  jj(fc{  +  V),  (4) 

in  which  kb  is  the  radius  of  gyration  of  the  base  (considered  as  a  thin  plate) 
with  respect  to  an  axis  through  its  center  of  gravity  parallel  to  the  axis 
through  the  vertex. 

4  6.  For  any  right  cylinder  or  prism,  show  that  the  radius  of  gyration  with 
respect  to  an  axis  perpendicular  to  the  length  through  the  center  of  one  end 
is  given  by  the  equation, 


7.  For  any  axis  perpendicular  to  the  length  of  any  right  cylinder  or  prism, 
show  that  the  radius  of  gyration  is  given  by  the  equation, 

k2  =  k\  +  kl, 

in  which  fci  is  the  radius  of  gyration  of  a  thin  rod  of  length  equal  to  that  of 
the  cylinder  or  prism  with  respect  to  a  perpendicular  axis,  and  kb  is  the 
radius  of  gyration  of  a  thin  plate  equivalent  to  a  section  of  the  cylinder  or 
prism  parallel  to  the  base. 

8.  By  means  of  Problem  7,  find  the  moment  of  inertia  of  a  homogeneous 
cylinder    10  inches  long,  4  inches  in  diameter,  and  weighing  24  pounds, 
with  respect  to  an  axis  perpendicular  to  its  length  and  tangent  to  one  end. 

Ans.     7  =  920. 


238  MECHANICS  [ART.  142 

9.  A  solid  cylinder,  10  inches  in  diameter  and  12  inches  long,  has  a  cylin- 
drical hole,  6  inches  in  diameter  arid  8  inches  long,  drilled  in  the  upper  end. 
The  axis  of  the  hole  is  parallel  to  the  axis  of  the  cylinder  and  1  inch  from  it. 
Find  the  square  of  the  radius  of  gyration  of  the  remainder  with  respect  to 
an  axis  in  the  upper  end  through  the  axis  of  the  hole  and  perpendicular  to 
the  plane  of  the  two  axes. 

141.  Moment  of  Inertia  of  Connected  Bodies. — To  find  the 
moment  of  inertia  of  two  or  more  connected  bodies  with  respect 
to  a  single  axis,  it  is  necessary  to  find  the  moment  of  inertia 
of  each  body  with  respect  to  a  parallel  axis  through  its  center 
of  gravity  and  then  transfer  to  the  common  axis.     The  moment 
of  inertia  of  all  the  bodies  with  respect  to  this  axis  is  found  by 
adding  the  moment  of  inertia  of  each  of  the  separate  parts. 
Since  moment  of  inertia  involves  the  square  of  a  distance,  it  is 
always    positive. 

Problems 

1.  A  solid  sphere  is  4  inches  in  diameter  and  weighs  10  pounds.     It  is 
hung  on  the  end  of  a  rod  1  inch  in  diameter  and  40  inches  long,  which  weighs 
9  pounds.     The  end  of  the  rod  is  tangent  to  the  surface  of  the  sphere.     Find 
the  radius  of  gyration  with  respect  to  a  diameter  of  the  rod  at  the  end  oppo- 
site the  sphere.  Ans.     k  =  34.37  in. 

2.  What  change  will  be  made  in  the  result  of  Problem  1  if  the  diameter 
of  the  rod  is  neglected? 

3.  A  right  cone,  4  inches  in  diameter  and  12  inches  high,  stands  on  the 
end  of  a  cylinder,  2  inches  in  diameter  and  20  inches  long.     The  axis  of  the 
cone  coincides  with  the  axis  of  the  cylinder.     The  cone  and  cylinder  have 
the  same  density.     Find  the  radius  of  gyration  with  respect  to  a    diameter 
of  the  cylinder  at  the  end  opposite  the  cone.  Ans.     17.77  in. 

142.  Summary. — Moment  of  inertia  is  defined  mathematically 
by  the  equations, 

/  =  fr*dm;  I  =  Zmrjj  Formula  XI 

in  which  /  is  the  moment  of  inertia/W  is  a  finite  element  of  mass, 
dm  is  an  infinitesimal  element  of  maJl,  and  r  is  the  distance  of  the 
element  from  the  axis  wifrT  res Jbt  to  which  the  moment  of 
inertia  is  taken. 

The  radius  of  gyration  is  cpfined  mathematically  by  the 
equation 

k2  =  -,  Formula  XIII 

m 

in  which  k  is  the  radius  of  gyration  and  m  is  the  entire  mass 
of  the  body.  The  radius  of  gyration  is  the  distance  from  the 


CHAP.  XIV]     MOMENT  OF  INERTIA  OF  SOLIDS          239 

axis  at  which  all  the  material  of  the  body  must  be  placed  in  order 
to  have  a  moment  of  inertia  equal  to  that  of  the  actual  body. 
When  the  moment  of  inertia  of  a  sphere  of  radius  a  is  taken  with 
respect  to  a  diameter, 

]fc«-^!. 
5 

When  the  moment  of  inertia  of  a  solid  cylinder  of  radius  a 
is  taken  with  respect  to  the  axis  of  the  cylinder, 

a2 

l/m  2         . 

"  2 

When  the  moment  of  inertia  of  a  right  cone  is  taken  with 
respect  to  the  axis  of  the  cone, 

*•  -  3°2 
w 

When  the  moment  of  inertia  of  a  rectangular  parallelepiped 
is  taken  with  respect  to  an  axis  through  the  center  perpendicular 
to  the  edges  of  length  b  and  d, 

fc2  +  d2 


k2  = 


12 


When  the  moment  of  inertia  of  a  right  cylinder  is  taken  for  an 
axis  through  the  center  perpendicular  to  its  length, 


For  any  right  cylinder  or  prism,  for  which  the  moment  of 
inertia  is  taken  with  respect  to  an  axis  perpendicular  to  its  length, 
the  square  of  the  radius  of  gyration  is  equal  to  the  square  of 
the  radius  of  gyration  regarded  as  a  thin  rod  plus  the  square  of 
the  radius  of  gyration  regarded  as  a  thin  plate. 

To  transfer  from  an  axis  through  the  center  of  gravity  to  a 
parallel  axis, 

jfcz  =  kl  +  d\  Formula  XV 

in  which  &0  is  the  radius  of  gyration  with  respect  to  the  axis 
through  the  center  of  gravity  and  d  is  the  distance  between  the 
axes. 

143.  Miscellaneous  Problems 

1.  A  homogeneous  hollow  cylinder  has  an  inside  diameter  of  6  inches  and 
and  outside  diameter  of  10  inches.  Find  its  radius  of  gyration  with  respect 
to  an  axis  12  inches  from  its  axis  and  parallel  to  it. 

Ans,     k  =  13.34  in. 


240  MECHANICS  [ART.  143 

2.  A  solid  of  revolution  is  formed  by  revolving  the  area  enclosed  by  the 
X  axis,  the  curve  y2  =  2x,  and  the  line  x  =  8  about  the  X  axis.     Find  the 
radius  of  gyration  of  this  solid  with  respect  to  the  X  axis. 

Ans.     k  =  2.309  in. 

3.  Find  the  radius  of  gyration  of  the  solid  of  Problem  2  with  respect  to 
the  Y  axis. 

4.  A  homogeneous  solid  cylinder,  2  inches  in  diameter  and  20  inches  long, 
is  attached  to  a  homogeneous  solid  hemisphere  8  inches  in  diameter.     The 
center  of  the  hemisphere  coincides  with  the  center  of  one  end  of  the  cylinder. 
If  the  density  of  the  hemisphere  is  the  same  as  the  density  of  the  cylinder, 
find  the  radius  of  gyration  with  respect  to  a  diameter  of  the  cylinder  at  the 
end  opposite  the  hemisphere. 

6.  Solve  Problem  4  if  the  density  of  the  hemisphere  is  twice  that  of  the 
cylinder. 


CHAPTER  XV 
MOMENT  OF  INERTIA  OF  A  PLANE  AREA 

144.  Definition. — In  calculating  the  strength  and  stiffness  of 
members  of  structures  and  machines,  much  use  is  made  of  the 
expression  Jr2dA,  in  which  dA  is  an  element  of  the  area  of  a 
plane  cross-section  of  the  member,  and  r  is  the  distance  of  the 
element  from  an  axis  in  the  plane  or  from  an  axis  normal  to  the 
plane. 

When  the  axis  is  normal  to  the  plane,  the  integral  is  equivalent 
to  the  moment  of  inertia  of  a  prism  of  cross-section  equal  to 
that  of  the  member  and  of  such  length  and  density  that  its  mass 
per  unit  area  is  unity.  When  the  axis  is  in  the  plane  of  the  sec- 
tion, the  integral  is  equivalent  to  the  moment  of  inertia  of  a 
plate  so  thin  that  the  square  of  its  thickness  is  negligible  and  of 
such  density  that  its  mass  per  unit  area  is  unity.  On  account  of 
the  mathematical  similarity  with 

/  =  fr2dm, 

the  expression  J  r2dA  is  called  the  moment  of  inertia  of  the  area. 

Considered  physically,  the  so-called  moment  of  inertia  of  an 
area  is  entirely  different  from  the  moment  of  inertia  of  a  solid. 
Considered  mathematically,  they  are  so  much  alike  that  it  is 
worth  while  to  treat  them  together.  A  formula  which  applies 
to  a  solid  of  uniform  thickness  may  also  be  applied  to  the  moment 
of  inertia  of  a  plane  area  by  the  substitution  of  area  for  mass. 

The  radius  of  gyration  of  a  plane  area  is  defined  by  the  equation 

79       moment  of  inertia  ,  . 

k2  = (1) 

area 

The  formula  for  the  transfer  of  axis  for  the  moment  of  inertia 
of  a  plane  area  is 

/  =  /o  +  Ad\  (2) 

145.  Polar  Moment  of  Inertia. — When  the  axis  is  normal  to  the 
plane  of  the  area,  JV°cL4.  is  called  the  polar  moment  of  inertia 
of  the  area.     The  polar  moment  of  inertia  of  a  plane  area  is 

16  241 


242 


MECHANICS 


[ART.    146 


equivalent  to  the  moment  of  inertia  of  a  right  prism  or  cylinder 
for  an  axis  parallel  to  its  length,  provided  the  density  and  length 
are  such  that  the  mass  per  unit  area  is  unity. 

It  is  customary  to  represent  the  polar  moment  of  inertia  by  J. 

J  =  fr'-dA.  Formula  XV 

In  Fig.  217,  0  represents  an  axis  perpendicular  to  the  plane  of  the 
paper.  The  element  of  area  is  rdOdr  and 

J  =  ffr*dOdr.  (1) 

In  Fig.  218,  the  element  of  area  is  dxdy  and 

J  =  fr*dA  =  //(z2  +  y^dxdy.  (2) 


dx 


FIG.  217. 


FIG.  218. 


Problems 


1.  Find  the  polar  moment  of  inertia  of  a  circle  of  radius  a  with  respect 
to  an  axis  through  its  center  and  compare  the  moment  of  inertia  and  radius 
of  gyration  with  those  of  a  cylinder. 

,       Tra*      Aa*    79       a2 
v  Ans.     J  =  —  =  — ;  fc2  ==  -. 

2.  Find  the  polar  moment  of  inertia  of  a  rectangle  of  breadth  b  and  depth 
d  with  respect  to  an  axis  through  its  center. 

Ans.     J  =     —^-  — 5  fc2       —\2~' 

3.  Find  the  polar  moment  of  inertia  of  an  isosceles  triangle  of  base  b 
and  altitude  h  with  respect  to  an  axis  through  the  vertex. 

,       bhz   .  Vh 

Ans.       J   =  —r-  +  27T* 

146.  Axis  in  Plane. — When  Jr*dA  is  taken  with  respect  to 
an  axis  in  the  plane  of  the  area,  it  is  called  simply  the  moment 
of  inertia  of  the  area.  This  moment  of  inertia  is  a  factor  in  the 
strength  and  stiffness  of  beams  and  columns,  and  has  much 
greater  application  than  the  polar  moment  of  inertia.  The 
values  of  this  moment  of  inertia  for  the  common  geometrical 
figures  and  for  sections  of  the  ordinary  structural  shapes  are  given 
in  the  handbooks  of  the  steel  companies. 


CHAP.  XV]       INERTIA  OF  A  PLANE  AREA 


243 


If  the  moment  of  inertia  is  taken  with  respect  to  the  X  axis, 
r  =  y,  and 

/  =  fy*dA  =  Ix.  (1) 

If  the  moment  of  inertia  is  taken  with  respect  to  the  Y  axis, 
r  =  x,  and 

/  =  fx*dA  =  Iy.  (2) 

For  convenience,  the  moment  of  inertia  with  respect  to  the 
X  axis  may  be  designated  by  I3  and  the 
moment  of  inertia  with  respect  to  the 
Y  axis  by  Iy.  (This  convention  is  con- 
venient. It  is  not,  however,  in  general 
use.) 

Example 

Find  the  moment  of  inertia  of  a  circular 
area  of  radius  a  with  respect  to  a  diameter. 

Figure  219  applies  to  this  example.  The 
moment  of  inertia  is  taken  with  respect  to 
the  -X"  axis  and  the  element  of  area  is  given 
in  polar  coordinates. 


FIG.  219. 


7  =  fy*dA  =  JJr3  sin2  edddr  =  |~^1   Jsin2  BdO', 


a4T0 


2w 


Problems 

1.  Find  the  radius  of  gyration  of  a  circular  area  with  respect  to  a  diameter. 

Ans.     k  =  — 

•r   2.  Find  the  moment  of  inertia  of  a  rectangle  of  base  b  and  altitude  d 
with  respect  to  a  line  through  the  center  parallel  to  the  base. 


Ans. 


I  =        . 

12 


3.  Find  the  moment  of  inertia  of  the  rectangle  of  Problem  2  with  respect 
to  the  base.     Solve  by  integration  and  also  by  transfer  of  axis. 

,       bd* 

Ans.     I  =  — • 
o 

4.  The  base  of  a  rectangle  is  6  inches  and  its  altitude  is  10  inches.     Find 
its  moment  of  inertia  with  respect  to  an  axis  4  inches  below  the  base  and 
parallel  to  it.     Solve  two  ways. 

Ans.     7  =  6  X^°00  +  60  X  81  =  5360  in.<; 


12 

6  X  2744 


6  X_64 
3 


=  5360  in.4 


244  MECHANICS  [ART.  146 

6.  By  integration  find  the  moment  of  inertia  of  a  triangle  of  base  6  and 
altitude  h  with  respect  to  an  axis  through  the  vertex  parallel  to  the  base. 

T       bh3 

Ans.     I  =  —r- 
4 

6.  Using  the  answer  of  Problem  5,  find  the  moment  of  inertia  of  a  triangle 
with  respect  to  an  axis  through  the  center  of  gravity  parallel  to  the  base, 
and  find  the  moment  of  inertia  with  respect  to  the  base. 

bh3     T      bh3 
Ans.     /„=_;/  =  _ 

7.  Find  the  moment  of  inertia  of  a  triangle  with  respect  to  the  base  by 
subtracting  the  moment  of  inertia  of  an  inverted  triangle  from  the  moment 
of  inertia  of  a  parallelogram. 

8.  A  trapezoid  has  a  lower  base  of  12 

-/  inches,  an  upper  base  of  6  inches,  and  an 

|  altitude  of  9  inches.     Find  its  moment  of 

J  inertia    with   respect   to  the  lower  base. 

I  Solve  by  means  of  two  triangles  and  check 


by  means  of  a  triangle  and  a  rectangle. 
Ans.     I  =  1822.5  in.4 


9.  Find    the    moment   of    inertia  of   a 
,,     6-inch  by  5-inch  by  1-inch  angle  section 
I        j  _  _  |_£      with  respect  to  an  axis  through  the  center 
of  gravity  parallel  to  the  5-inch  leg.    (Fig. 
220.) 

The  center  of  gravity  is  found  to  be  2 
inches  from  the  back  of  the  5-inch  leg. 

To  find  the  moment  of  inertia  with  respect  to  the  axis  00',  the  sec- 
tion is  divided  into  a  6-inch  by  1-inch  rectangle  and  a  1-inch  by  4-inch 
rectangle.  The  moment  of  inertia  of  each  rectangle  is  found  with  respect 
to  a  vertical  line  through  its  center  of  gravity,  and  then  transferred  to  the 
axis  00'. 

The  moment  of  inertia  of  the  6-inch  by  1-inch  rectangle  with  respect  to 
the  vertical  line  through  its  center  of  gravity  is  18  in.4  This  axis  is  1  inch 
from  the  axis  00'.  The  moment  of  inertia  of  the  6-inch  by  1-inch  rectangle 
with  respect  to  00'  is  18  +  6  X  I2  =  24  in.4  The  moment  of  inertia  of 
the  1-inch  by  4-inch  rectangle  with  respect  to  OO'  is  0.33  +  4  X  1.52  = 
9.33.  The  moment  of  inertia  of  the  entire  section  with  respect  to  00'  is 
24  +  9.33  =  33.33  in.4 

The  problem  may  also  be  solved  by  finding  first  the  moment  of  inertia 
of  the  entire  section  with  respect  to  some  vertical  axis  and  then  by  transfer- 
ring to  the  parallel  axis  through  the  center  of  gravity.  The  moment  of  inertia 
with  respect  to  the  back  of  the  5-inch  leg  is 

_  1  X6*       4X1*  _  „. 


This  axis  is  2  inches  from  the  axis  00'. 

70  =  731  -  10  X  22  =  331. 
10.  Solve  Problem  9  by  dividing  the  section  into  two  equal  rectangles 


CHAP.  XV]       INERTIA  OF  A  PLANE  AREA 


245 


as  shown  in  Fig.  221.     For  the  second  method,  find  I  with  respect  to  the 
axis  CC'  and  then  transfer  to  00'  through  the  center  of  gravity. 

11.  Find  the  moment  of  inertia  of  the  6-inch  by  5-inch  by  1-inch  angle 
section  of  Fig.  220  with  respect  to  the  axis  through  the  center  of  gravity 
parallel  to  the  6-inch  leg.     Calculate  the  radius  of  gyration  and  compare 
the  results  with  those  of  Cambria  Steel  or  Carnegie  Pocket  Companion 
for  a  similar  smaller  section. 

12.  Figure  222  shows  a  standard  channel 
section  without  fillets  or  curves.     The  slope 

of  the  flange  of  standard  section  is  «•    Find 

the  moment  of  inertia  of  this  section  with 
respect  to  an  axis  through  the  center  of 
gravity  perpendicular  to  the  web. 

M3    h*  - l* 

12   "        16 


Ans.     I  =  — 


C'   0' 

1 

1 

/ 

tr 

1 

1 

1 
1 

1 

1 

*•/, 
-i. 

1    1 

c    o 

FIG.  221. 

13.  Take  the  dimensions  from  a  steel  hand- 
book and  calculate  the  moment  of  inertia  of 
a      12-inch     20.5-pound     channel     section. 
Compute  the  area  of  the  section  and  the 
radius  of  gyration. 

14.  With  the  moment  of  inertia  of  a  12-inch  20.5-pound  channel  known, 
calculate  the  moment  of  inertia  of  a  12-inch  30-pound  channel  and  a  12-inch 
40-pound  channel. 

15.  Find  the  distance  of  the  center  of  gravity  of  a  12-inch  20.5-pound 
channel  from  the  back  of  the  web.     Find  the  moment  of  inertia  with  respect 

to  the  back  of  the  web 
and  transfer  to  the  par- 
/  allel  axis  through  the 
center  of  gravity.  Com- 
pare results  with  the 
handbook. 

16.  Look    up    the    ex- 
pression for  the  moment 
of  inertia  of  an  I-beam 
section  for  axis    perpen- 
dicular to  the  web.     De- 
rive the  expression.     Do 
the    same    for    the   axis 
through   the  center  par- 
allel to  the  web. 

17.  Find  the  moment 
of  inertia  and  the  radius 
of  gyration  of  a  15-inch 

42-pound  I-beam  for  axes  through  the  center  of  gravity  perpendicular  to 
the  web  and  parallel  to  the  web.  Compare  the  results  with  the  handbook. 
18.  Figure  223  gives  the  dimensions  of  a  plate-and-angle  section  made 
up  of  one  14-inch  by  1-inch  plate  and  four  6-inch  by  3$-mch  by  1-inch 
angles.  Look  up  the  moment  of  inertia  and  the  location  of  the  center  of 
gravity  of  one  of  these  angles  in  the  handbook  and  compute  the  moment 


FIG.  222. 


FIG.  223. 


246 


MECHANICS 


[ART.    147 


of  inertia  of  the  section  with  respect  to  an  axis  through  the  center  of  the 
section  perpendicular  to  the  plate  and  for  an  axis  parallel  to  the  plate. 
Calculate  the  radius  of  gyration  and  compare  with  the  handbook. 

19.  Two  10-inch  15-pound  channels  are  placed  vertical  and  riveted  to  a 
12-inch  by  ^-inch  plate  at  the  top.     Find  the  moment  of  inertia  with  respect 
to  an  axis  through  the  center  of  gravity  perpendicular  to  the  channels. 
Use  all  the  data  you  can  get  from  the  handbook.     Where  is  such  a  section 
used? 

20.  A  circle  of  radius  a  has  a  segment  cut  off  by  means  of  a  chord  at  a 

distance  ^  from  the  center.     By  integration  find  the  moment  of  inertia  of 

the  area  between  the  chord  and  the  circle  with  respect  to  the  chord  as  an  axis. 

21.  Find  the  moment  of  inertia  of  the  area  between  the  curves  yz  =  Qx 
and  z*  =  Qy  with  respect  to  the  X  axis. 

22.  Find  the  moment  of  inertia  of  the  area  bounded  by  the  curve  xy  =  24, 
the  line  x  =  2,  and  the  line  y  —  3  with  respect  to  the  X  axis. 

23.  Find  the  moment  of  inertia  of  the  area  bounded  by  the  curve  y  =  xz, 
the  line  x  =  4  and  the  X  axis  with  respect  to  the  line  x  =  —  2. 

147.  Relation  of  Moments  of  Inertia.  —  If  dA  is  an  element  of 
area  in  the  XY  plane,  the  moment  of  inertia  of  the  entire  area 
with  respect  to  the  X  axis  is 

J,  =  J>dA.  (1) 

With  respect  to  the  Y  axis  +^* 


I    = 


(2) 


The  polar  moment  of  inertia  with  respect  to  an  axis  through  the 
origin  is 

c  c  c  c 

The  sum  of  the  moments  of  inertia  of  a  plane  area  with  respect 
to  two  axes  in  its  plane  which  are  perpendicular  to  each  other  is 

^,, equal   to   the   polar  moment  of  inertia 

of  the  area  with  respect  to  the  inter- 
section of  these  axes.  It  follows  that 
the  sum  of  the  moments  of  inertia 
of  a  plane  area  with  respect  to 
any  pair  of  axes  in  its  plane  which 
are  at  right  angles  to  each  other 
\N  and  pass  through  the  same  point 

FIG.  224.  is  constant. 

Example 

Find  the  moment  of  inertia  of  a  4-inch  by  3-inch  rectangle  with  respect 
to  a  fine  through  one  corner  perpendicular  to  the  diagonal. 

The  diagonal  OC  of  Fig.  224  divides  the  rectangle  into  two  triangles, 


/ 


CHAP.  XV]        INERTIA  OF  A  PLANE  AREA 


247 


each  of  which  has  a  base  of  5  inches  and  an  altitude  of  2.4  inches.  The 
moment  of  inertia  of  each  triangle  with  respect  to  OC  is  5.76  in.4  and  the 
moment  of  inertia  of  the  rectangle  is  11.52  in.4  If  moments  of  inertia  are 
taken  with  respect  to  horizontal  and  vertical  axes  through  0, 

Ix  =  36;  /„  =  64; 
I*  +  Iy  =  J  =  11.52  +  7; 
/  =  88.48  in.4 

Problems 

1.  The  moment  of  inertia  of  a  circle  with  respect  to  a  diameter  is  — j-  • 
By  the  principle  of  Equation  (3)  show  that  the  polar  moment  of  inertia 
with  respect  to  the  center  is  —~-- 

2.  The  polar  moment  of  inertia  of  a  circle  with  respect  to  a  point  at  its 

37T04 

2 


circumference  is 


Find  the  moment  of  inertia  of  a  circle  with  respect 


to  a  tangent  in  its  plane.     Check  by  the  transfer  of  axis. 

3.  Find  the  moment  of  inertia  of  a  6-inch  by  6-inch  by  1-inch  angle 
section  with  respect  to  the  axis  which  bisects  the  angle  between  the  legs  of 
the  section.  Use  Equation  (3)  and  get  all  the  data  from  the  handbook. 
Check  by  completing  the  square  and  subtracting  the  moment  of  inertia  of 
the  additional  area  from  the  moment  of  inertia  of  the  square. 

148.  Product  of  Inertia. — To  find  the  moment  of  inertia  of  a 
plane  figure  with  respect  to  some  inclined  axis  in  its  plane  often 
involves  difficult  integrations  or  limits. 
These  difficulties  may  be  avoided  by 
finding  the  moment  of   inertia  with 
respect  to  a  pair  of  axes  for  which  the 
calculations  are  easiest  and  then  trans- 
ferring to  a  new  axis  at  the  required 
angle  with  these  axes.     In  making  this 
transformation  (Art.  150)  one  of  the 
terms  is  found  to  be   \  xydA .    This  in- 
tegral is  called  the  product  of  inertia  of 
the  area.    The  product  of  inertia  is  rep- 
resented by  H  in  algebraic  equations. 

H  =  fxydA 


FIG.  225. 


(1) 


The  product  of  inertia  has  no  physical  significance.  It  is,  how- 
ever, so  useful  in  the  calculation  of  the  moment  of  inertia  of 
plane  areas  that  it  is  desirable  to  master  some  of  its  properties. 


248  MECHANICS  [ART.  149 

//  an  area  has  an  axis  of  symmetry,  the  product  of  inertia  with 
respect  to  that  axis  and  the  axis  perpendicular  to  it  is  zero.  In 
Fig.  225,  the  Y  axis  is  an  axis  of  symmetry.  Any  horizontal 
line  extends  equal  distances  on  each  side  of  the  Y  axis.  If 
Equation  (1)  is  integrated  first  with  respect  to  x 

H  =  fxydA  =  ffxydydx  =  -[z2]   ydy  =  £/|*i  r  ^i]  ydy.- 

When  the  area  is  symmetrical  with  respect  to  the  Y  axis,  x\  is 
numerically  equal 'to  xz  but  has  the  opposite  sign.  Since  the 
square  of  a  negative  quantity  is  positive,  x\  has  the  same  sign 
as  x\  and  x\  —  x\  =  0.  Consequently 

H  =  0.  (2) 

Problems 

1.  By  integration  find  the  product  of  inertia  of  a  rectangle  of  width  b 
and  height  d  with  respect  to  the  lower  edge  and  the  left  edge  as  axes. 

Ans.     H  =  —r-' 
4 

2.  Solve  Problem  1  with  respect  to  the  lower  edge  and  the  right  edge  as 
axes.  Ans.     H  = j-* 

3.  Find  the  product  of  inertia  of  the  first  quadrant  of  a  circle  of  radius 
a  with  respect  to  the  X  and  Y  axes.  Ans.     H  =  -^> 

o 

4.  Find  the  product  of  inertia  with  respect  to  the  coordinate  axes  of  the 
area  bounded  by  these  axes  and  the  line  whose  intercepts  are  x  =  6  and 

y  =  4.  A?is.     H  =  24  in.4 

6.  Find  the  product  of  inertia  with  respect  to  the  coordinate  axes  of  the 
area  bounded  by  these  axes  and  the  line  whose  intercepts  are  (0,  d)  and 
(b,  0). 

149.  Transfer  of  Axes  for  Product  of  Inertia.— In  Fig.  226, 
OX  and  OF  are  a  pair  of  axes  at  right  angles  to  each  other,  and 
O'X'  and  O'Y'  are  a  second  pair  of  axes  parallel  to  these.  The 
coordinates  of  the  point  0  with  respect  to  the  axes  O'X'  and  0'  Y' 
are  a  and  b.  It  is  required  to  find  the  product  of  inertia  of  the 
area  with  respect  to  O'X'  and  O'Y'. 

The  coordinates  of  an  element  dA  with  respect  to  OX  and  0  Y 
are  x  and  y.  With  respect  to  O'X'  and  O'Y',  the  coordinates 


CHAP.  XV]       INERTIA  OF  A  PLANE  AREA 


249 


are  a  -f-  x  and  b  +  y. 
equation, 

H 


The  product  of  inertia  is  given  by  the 


f(a  +  x)(b  +  y)dA  =  abfdA  +  afydA  + 

bfxdA  +  fxydA.     (1) 

The  first  term,  dbfdA,  is  equivalent 
to  abA.  The  last  term  fxydA.  is 
HQ,  which  is  the  product  of  inertia 
with  respect  to  the  axes  OX  and  OF. 
The  term  fydA.  is  the  moment  of 
the  area  with  respect  to  the  X  axis  and 
the  term  fxdA  is  the  moment  of  the 
area  with  respect  to  the  Y  axis.  If  o1 
the  point  0  is  at  the  center  of  grav- 
ity of  the  area,  these 
and 


FIG.  226. 


Formula  XVI 


If  either  OX  or  0  F*1s--aft~ftxi&_of  symnigfer^  the  last  term  also 
vanishes.  While  the  moment  of  inertia  is  always  positive,  the 
product  of  inertia  may  be  either  positive  or  negative. 

Problems 

1.  Find  the  product  of  inertia  of  a  rectangle  6  inches  wide  and  4  inches 
high  with  respect  to  the  lower  edge  and  the  left  edge  as  axes. 

Ans.     H  =  3  X  2  X  24  =  144  in.4 

2.  Find  the  product  of  inertia  of  the  rectangle  of  Problem  1  with  respect 
to  the  lower  edge  and  the  right  edge  as  axes. 

Ans.     H  =  (-3)2  X  24  =  -144  in.4 

3.  Find  the  product  of  inertia  of  a  semi- 
circle in  the  first  and  fourth  quadrants  with 
respect  to  the  diameter  which  bounds  it  and 
the  tangent  at  the  lower  end  of  this  diameter. 

Ans.     H  =  ^. 

4.  Find  the  product  of  inertia  of  a  6-inch 
by   5-inch    by   1-inch   angle   section  in   the 
position  of  Fig.  227  with  respect  to  the  back 
of  the  legs  as  axes. 

5.  Find  the  product  of  inertia  of  the  angle 
section  of  Problem  4  with  respect  to  axes  through  the  center  of  gravity 
parallel  to  the  legs. 

The  section  may  be  divided  into  two  rectangles.  The  product  of  inertia 
may  be  transferred  from  lines  through  the  center  of  gravity  of  each  of  these 
rectangles  to  the  parallel  lines  through  the  center  of  gravity  of  the  figure. 


Y 

*\ 

5"    1 

1 
i 

2-> 

0 

-  -4— 

\r 

—  TTTT*  —  * 

£ 

hy'^" 

4L      • 

^ 

L 

«             j 

FIG.  227. 


250 


MECHANICS 


[ART.   151 


Another  way  is  to  take  the  product  of  inertia  with  respect  to  the  axes  O'X' 
and  O'Y'  which  bisect  the  legs.  The  product  of  inertia  for  these  axes  is 
zero,  since  O'X'  is  an  axis  of  symmetry  for  the  horizontal  leg  and  O'Y'  is 
an  axis  of  symmetry  for  the  vertical  leg.  Transferring  to  the  center  of 
gravity, 

0  =  Ho  +  abA  =  H0  +  1.5  X  1  X  10 
#o  =  -15  in.4 


150.  Change  of  Direction  of  Axis  for  Moment  of  Inertia.  —  In 

Fig.  228,  OX  and  OY  are 
a  pair  of  axes  for  which 
the  moment  of  inertia  is 
known.  These  are  Ix  = 
fyfdA  and  Iy  =  fx^dA  . 
It  is  desired  to  find  the 
moment  of  inertia  with 
respect  to  the  axis  OX' 
which  makes  an  angle  6 
with  OX. 

Let  OY'  be  an  axis 
.perpendicular  to  OX'. 
The  coordinates  of  the  element  dA  with  respect  to  OX'  and 
OY'  as  axes  are  (x',  y').  The  moment  of  inertia  with  respect 
to  OX'  is 

/'  =  Sy'*dA.  (1) 


From  the  geometry  of  the  figure, 

y'  =  y  cos  0'  —  x  sin  0; 

/'  =  f(y*  cos2  6  -  2xy  cos  B  sin  0  +  x2  sin2 
/'  =  Ix  cos2  e  +  Iy  sin2  6  -  2  cos  0  sin  BfxydA ; 
/'  =  Ix  cos2  0  +  Iy  sin2  0  -  H  sin  20; 

/'  =  ^4^  +  IjL^  cos  20  -  H  sin  20. 


(2) 
(3) 
(4) 
(5) 

(6) 


Problems 

1.  Find  the  moment  of  inertia  of  a  6-inch  by  4-inch  rectangle  with  respect 
to  an  axis  through  the  center  at  an  angle  of  25  degrees  with  the  6-inch  edges. 

Ix  =  32,  /„  =  72,  H  =  0; 

/'  =  52  -  20  cos  50°  =  52  -  20  X  0.6428  =  39.144  in." 

2.  Find  the  moment  of  inertia  of  the  rectangle  of  Problem  1  with  respect 
to  the  diagonal  by  means  of  Equation  (6).     Check  by  dividing  the  rectangle 
into  two  triangles  and  finding  the  moment  of  inertia  of  each  with  respect 
to  the  diagonal  as  the  base. 

3.  Find  the  moment  of  inertia  of  the  rectangle  of  Problem  1  with  respect 


CHAP.  XV]       INERTIA  OF  A  PLANE  AREA 


251 


to  an  axis  through  one  corner.     The  axis  makes  an  angle  of  20  degrees  with 
the  6-inch  edge,  and  cuts  one  4-inch  edge. 

Ans.     I'  =  208  -  61.28  -  92.56  =  54.16  in.4 

151.  Change  of  Direction  of  Axes  for  Product  of  Inertia. — 

From  Fig.  228,  the  product  of  inertia  with  respect  to  the  axes 
OX'  and  OY'  is 

H'  =  fx'y'dA-,  (1) 

Hf  =  f(x  cos  6  +  y  sin  0)(y  cos  0  -  x  sin  0)  dA;  (2) 
H'  =  (cos2  6  -  sin2  6)fxydA  + 

cos  B  sin  0f(y2  -  x2)dA;  (3) 


H'  =  H  cos  26  + 


Ix       Iy  sin  29. 


(4) 


If  H'  is  zero  the  second  member  of  Equation  (4)  is  zero,  and 
H  cos  6  =     v  ~    x  sin  20;  (5) 

tan  20  =  T  2H  T-  (6) 


/,- 

— 

Since  the  tangent  of  an  angle  may  have  any  value  from  minus 
infinity  to  plus  infinity,  there  is  always  some  direction  for  which 
the  product  of  inertia  for  a  pair  of  axes  through  any  point  is 
zero.  If  there  is  a  line  of  symmetry  through  the  point,  this  line 
is  one  of  the  axes  for  which  the  product  of  inertia  is  zero. 

Problems 

1.  Find  the  direction  of  the  axes  through  the  lower  left  corner  of  a  rect- 
angle 6  inches  wide  and  4  inches  high  for 

which  the  product  of  inertia  is  zero.     (Fig. 
229.) 

Ans.     30°  28'  and  120°  28'  with 
the  6-inch  edge. 

2.  Find  the  direction  of  the  axes  through 
the  center  of  gravity  of  a  6-inch  by  4-inch 
by  1-inch  angle  section  for  which  the  pro- 
duct of  inertia  is  zero. 

Ans.     Tan  26  =  -1;  6  =  -22°  30 
'  and  67°  30'. 


<  ^  —  --> 

-3o-sa'      ^^ 

^:-> 

/' 

\ 

•^ 

\ 

s     ^ 

FIG.  229. 


152.  Maximum  Moment  of  Inertia.— To  find  the  direction  of 
the  axis  through  any  given  point  for  which  the  moment  of  inertia 
is  greater  than  for  any  other  axis  through  that  point,  the  mathe- 


252  MECHANICS  [ART.  153 

matical  conditions  for  maximum  and  minimum  are  applied  to 
Equation  (6)  of  Art.  150. 

7,  =  I*  +  Iy  +  I*_^JjLCOs20  -  Hsiu20.  (1) 

' 


Differentiating  with  respect  to  0, 

~  =  (/„-/*)  sin  2(9  -  2H  cos  20.  (2) 

a0 

Equating  the  derivative  to  zero  and  solving  for  0, 

2/7  S 

tan  20  =  ^r-  -y-  K  (3) 

1  y  1  x 

Equation  (3)  is  identical  with  Equation  (6)  of  Art.  151.  It  is 
evident,  therefore,  that  the  directions  for  which  the  moment 
of  inertia  is  a  maximum  or  a  minimum  coincide  with  the  two 
directions  for  which  the  product  of  inertia  is  zero. 

Two  angles  which  are  180  degrees  apart  have  the  same  tangent. 
Equation  (3),  therefore,  gives  two  values  of  0,  which  differ  by  90 
degrees.  For  one  of  these  directions,  the  moment  of  inertia  is 
a  maximum;  for  the  other,  the  moment  of  inertia  is  a  minimum. 

If  an  axis  of  symmetry  passes  through  a  given  point,  the 
moment  of  inertia  for  that  axis  is  either  a  maximum  or  a 
minimum. 

Problems 

1.  Find  the  maximum  and  minimum  moment  of  inertia  of  a  6-  inch  by 
4-  inch  rectangle  for  axes  through  the  lower  left  corner.     (Fig.  "229.) 

From  Problem  1  of  Art.  151,  tan  20  =  1.800;  20  =  60°  57'  and  240°  57'. 
/m»»  =  208  -  80  cos  60°  57'  -  144  sin  60°  57'  =  43.27  in.4 
Ima*  =  208  -  80  cos  240°  57'  -  144  sin  240°  57'  =  372.73  in." 
I  mi*  is  the  moment  of  inertia  with  respect  to  OM  of  Fig.  229. 

2.  Find  the  least  moment  of  inertia  and  the  least  radius  of  gyration  of 
a  6-inch  by  6-inch  by   1-inch  angle  section.     Take  Ix  and  Iv  from  the 
handbook. 

3.  Find  the  maximum  and  minimum  moments  of  inertia  for  axes  through 
the  center  of  gravity  of  a  6-inch  by  4-inch  by  1-inch  angle  section.     Take 
Ix  and  7y  from  handbook.     Calculate  the  least  radius  of  gyration  and 
compare  the  result  with  the  handbook. 

153.  Center  of  Pressure. — In  Art.  104,  the  center  of  pressure 
of  a  liquid  on  a  plane  surface  was  defined  as  the  point  of  applica- 
tion of  the  resultant  of  all  the  pressure  on  one  side  of  the  sur- 


CHAP.  XV]       INERTIA  OF  A  PLANE  AREA 


253 


face.     Only  the  simplest  area  could  be  conveniently  calculated 
by  the  methods  given  there. 

In  Fig.  230,  dA  is  an  element  of  a  vertical  surface  subjected 
to  liquid  pressure  on  one  side.  All  parts  of  the  element  are  at  a 
distance  y  below  the  surface  of  the  liquid.  If  w  is  the  weight  of 
the  liquid  per  unit  volume, 


ctA  — 


horizontal  pressure  on  dA  =  wydA. 

The  moment  of  the  pressure 
about  the  line  of  intersection 
of  the  plane  of  the  surface  with 
the  surface  of  the  liquid  is 

dM  =  wy*dA;  (2) 

M  =  wfyzdA.          (3) 

The  integral  of  Equation  (3) 
is  the  moment  of  inertia  of  the 
area  with  respect  to  the  line 
of  intersection  with  the  surface 
of  the  liquid. 

M  =  vjf.  (4) 

If  yc  is  the  distance  of  the  center 
of  pressure  from  the  surface  of 
the  liquid,  and  if  y  is  the  dis- 
tance of  the  center  of  gravity 
of  the  area  from  the  surface  of  FIG.  230. 

the  liquid,  then 

moment  wl     _  k2 

total  pressure       wyA         y' 


(1) 


(5) 


in  which  k  is  the  radius  of  gyration  of  the  area  with  respect 
to  the  line  of  intersection  of  its  plane  with  the  surface  of  the 
liquid 

Problems 

1.  Find  the  center  of  pressure  of  a  vertical  rectangular  gate  of  width 
b  and  height  h  when  the  liquid  reaches  the  top. 

7        bh*      bh2      2h 
Vc  =  $A  =  ~y  *  ~2   =  T' 

2.  A  vertical  rectangular  gate  is  6  feet  wide  and  4  feet  high  and  is  subjected 
to  the  pressure  of  water  which  rises  6  feet  above  the  top  of  the  gate.     Find 
the  center  of  pressure.     Ans.     8.167  ft.  below  the  surface  of  the  water. 


254 


MECHANICS 


[ART.    153 


If  fco  is  the  radius  of  gyration  of  the  surface  with  respect 
to  the  horizontal  axis  in  its  plane  through  the  center  of  gravity, 

fc2  =  fcj  +  y\  (6) 

and  Equation  (5)  becomes 


y*  = 


y  v 

The  distance  of  the  center  of  pressure  below  the  center  of  gravity 
of  the  area  is  equal  to  the  square  of  the  radius  of  gyration  of 
the  area  divided  by  the  distance  of  its  center  of  gravity  below  the 
surface  of  the  liquid. 

Problems 

'  3.  A  vertical  circular  gate  of  radius  r  is  subjected  to  the  pressure  of 
liquid  which  just  reaches  the  top.     Find  the  center  of  pressure. 


Ans. 


-  below  the  center  of  the  circle. 
4 


pressure 


4.  A  vertical  circular  gate,  4  feet  in  diameter,  is  subjected  to  the  pressure 
of  a  liquid  which  reaches  10  feet  above  the  center.     Locate  the  center  of 

Ans.     0.1  ft.  below  the  center. 

5.  A  vertical  circular  gate,  6 
feet  in  diameter,  is  subjected  to 
the  pressure  of  water  which  rises 
to  the  center.     The  gate  is  held 
by  two  bolts  at  opposite  ends  of 
the    horizontal   diameter  and  a 
single  bolt  at  the  bottom.     Find 
the  tension  in  each  bolt. 

Ans.  662.7  Ib.  in  the 
lower  bolt. 

6.  A  solid  cylinder,   1  foot  in 
diameter,  is  cut  by  two  planes. 
One  plane  is  perpendicular  to  the 
axis  and  the  other  plane  cuts  the 
cylinder  so  that  the  longest  ele- 
ment   is    2    feet    long    and   the 
shortest  element  is  1  foot  long. 
Use  the  equations  above  to  find 

the  center  of  gravity  of  the  volume  between  the  planes. 

Ans.     0.5  inch  from  the  axis. 

Figure  231  shows  liquid  pressure  on  an  inclined  surface.  The 
line  BD  is  the  intersection  of  the  plane  of  the  inclined  surface 
with  the  surface  of  the  liquid.  The  element  dA  is  at  a  distance 
y'  from  the  line  BD.  If  6  is  the  angle  which  the  plane  makes  with 
the  vertical,  the  vertical  distance  of  the  element  of  area  below 


FIG.  231. 


CHAP.  XV]       INERTIA  OF  A  PLANE  AREA 


255 


the  surface  of  the  liquid  is  y'  cos  0,  and  the  pressure  on  the  element 
is  wy'  cos  6dA.     The  moment  equation  about  BD  is 


M  =  w  cos  6fy'2dA  =  WF  cos  0. 


(8) 


in  which  /'  is  the  moment  of  inertia  of  the  area  with  respect 
to  the  axis  BD. 


wif  cos  e       r 

wy'A  cos  0        y'A 


(9) 


Equation  (9)  is  the  same  as  Equation  (5)  except  that  the  dis- 
tances are  not  vertical.  An  equation  may  be  written  which 
is  similar  to  Equation  (7). 

154.  Moment  of  Inertia  by  Moment  of  a  Mass. — In  the  pre- 
ceding articles  it  was  shown  that  the  moment  of  a  pressure  which 
varies  as  the  distance  from  the  axis  of  moments  may  be  calculated 
by  means  of  the  moment  of  inertia  of  the  plane  area.  From 
Equation  (5)  of  Art.  153 


wl  =  wyAyc; 
I  =  yAyc. 


(1) 

(2) 


From  Equation  (2),  the  moment  of  inertia  of  a  plane  area  with 
respect  to  a  given  axis  is  the  product  of  the  area  A,  multiplied 
by  the  distance  of  its  center  of  gravity  from  the  axis,  and  by  the 
distance  of  the  center  of  a  pressure  which  varies  as  the  distance 
from  the  axis.  The  center  of  pressure  is  located  at  the  center  of 
gravity  of  a  solid,  the  base  of  which  is  the  given  area  and  the  height 
is  proportional  to  the  distance  from  the  axis. 

The  center  of  gravity  of  a  solid  may  be  found  experimentally 
by  moments.  Figure  232  shows  a  convenient  method.  The 
body  whose  center  of  gravity  is  desired  is  placed  on  a  balance 
beam  and  carefully  balanced  with  the  beam  in  a  horizontal  posi- 
tion. It  is  then  turned  180  degrees  about  a  vertical  axis  to  the 
position  shown  by  the  broken  lines  of  the  figure,  and  moved  along 


256 


MECHANICS 


[ART.    154 


the  beam  until  exact  balance  is  obtained  with  no  change  of  the 
weights.  The  center  of  gravity  is  then  at  the  same  location  as  it 
was  in  the  first  position.  If  C  is  a  point  in  the  body,  and  C"  is  the 
same  point  after  the  body  is  turned,  the  center  of  gravity_is 
located  in  a  vertical  line  midway  between  C  and  C'. 

Figure  233  represents  a  section  of  a  beam  whose  moment  of 
inertia  is  desired  for  a  line  through  the  center  of  gravity  parallel 
to  a  given  line  00'.  If  the  section  is  large,  the  outlines  may  be 
traced  on  paper  and  the  area  determined  by  a  planimeter.  If 
the  section  is  small,  a  portion  of  the  beam  of  convenient  length 
may  be  cut  off  between  two  sections.  The  volume  of  this  por- 
tion may  be  determined  by  weighing  in  air  and  in  water  and  its 
area  may  then  be  calculated  by  dividing  the  volume  by  the 
length. 


FIG.  233. 


The  center  of  gravity  of  the  section  may  be  found  by  balancing 
this  portion  of  uniform  thickness  upon  the  beam  of  a  balance,  as 
in  Fig.  232.  The  line  CC",  Fig.  233,  passes  through  the  center  of 
gravity. 

To  get  the  moment  of  inertia,  a  force  is  required  which  varies 
directly  as  the  distance  from  the  center  of  gravity.  Wedge- 
shaped  pieces  are  cut  from  the  beam  by  planes  which  intersect 
at  AB.  The  plane  ABD  of  Fig.  233,  II,  is  normal  to  the  length 
of  the  beam,  and  the  plane  ABE  makes  a  convenient  angle  with 
the  length.  This  wedge-shaped  piece  is  balanced  as  in  Fig.  232 
and  its  center  of  gravity  determined.  The  distance  of  the  center 
of  gravity  from  the  line  AB  gives  Yc  for  this  part  of  the  section. 
The  location  of  the  center  of  gravity  of  the  part  to  the  right  of 
AB  is  found  by  balancing  a  piece  of  uniform  length.  This 
experiment  gives  y. 

The  product  y(Ay  for  the  part  to  the  right  of  AB  gives  the 
moment  of  inertia  of  that  part  of  the  section.  A  similar  set  of 


CHAP.  XV]        INERTIA  OF  A  PLANE  AREA  257 

experiments  gives  the  moment  of  inertia  of  the  part  of  .the  section 
to  the  left  of  A  B.  The  moment  of  inertia  of  the  entire  section  is 
the  sum  of  these  two  results. 

Problems 

1.  A  portion  of  a  beam,  0.25  inch  in  length,  weighs  0.353  Ib.  in  air  and 
0.309  Ib.  in  water.     Find  the  area  of  the  section.     Ans.     A  =  4.88  in.2 

2.  The  section  of  Problem  1  was  5  inches  wide  from  a  point  F  to  a  point 
D.     It  was  placed  on  a  balance  beam  with  F  to  the  left  and  balanced.     When 
it  was  turned  end  for  end  and  balanced  again,  it  was  found  that  the  point 
F  was  4.24  inches  from  the  position  of  the  previous  balance.     How  far  is 
the  center  of  gravity  from  F?  Ans.     =  2.12  in. 

3.  The  beam  of  Problems  1  and  2  was  cut  by  two  planes  which  intersected 
on  the  line  FH  of  Fig.  233,  I.     One  plane  was  normal  to  the  length  and  the 
other  made  an  .angle  with  it.     The  portion  was  placed  on  the  scale  beam  with 
the  normal  section  horizontal  and  it  was  found  that  the  center  of  gravity 
of  the  wedge  was  3.21  inches  from  FH.     Find  the  moment  of  inertia  of  the 
section  with  respect  to  the  axis  FH  and  with  respect  to  the  parallel  axis 
through  the  center  of  gravity  of  the  section. 

Ans.     I  =  33.22    in.*;    /„  =  11.29    in.4 

155.  Summary.  —  The  moment  of  inertia  of  a  plane  area  is 
defined  mathematically  by  the  expression, 

Moment  of  inertia  = 


When  the  axis  is  perpendicular  to  the  plane  of  the  area,  the 
moment  of  inertia  is  called  the  polar  moment  of  inertia.  Polar 
moment  of  inertia  is  represented  by  the  letter  J.  The  moment  of 
inertia  of  a  plane  area  with  respect  to  an  axis  in  its  plane  is 
represented  by  the  letter  /. 

The  sum  of  two  moments  of  inertia  with  respect  to  axes  in 
the  plane  of  the  area  at  right  angles  to  each  other  is  equal  to  the 
polar  moment  of  inertia  with  respect  to  an  axis  through  the 
intersection  of  these  two  axes. 

J  =  h  +  /,. 
The  formulas  for  the  transfer  to  parallel  axes  are 

7  =  70  +  Ad2;J  =  Jo  +  Ad2. 
Radius  of  gyration  is  given  by 

/c2  =  4-  °r  fc2  =  T- 
A  A 

Product  of  inertia  is  defined  by  the  equation 
H  =  fxydA. 

17 


258  MECHANICS  [ART.   155 

Product  of  inertia  is  useful  in  changing  the  direction  of  the  axis 
for  moment  of  inertia,  and  in  finding  the  direction  of  the  axes 
for  which  the  moment  of  inertia  is  a  maximum  or  a  minimum. 
Moment  of  inertia  is  a  maximum  for  one  of  the  pair  of  axes 
for  which  the  product  of  inertia  is  zero  and  a  minimum  for  the 
other.  Product  of  inertia  is  zero  if  either  axis  is  an  axis  of  sym- 
metry. Moment  of  inertia  is  a  maximum  or  minimum  and  prod- 
uct of  inertia  is  zero  for  the  directions  which  satisfy  the  equation 

2H 


tan  26 


I,  -  L 


The  location  of  the  center  of  pressure  of  a  liquid  on  a  plane 
surface  is  given  by  the  equations, 


in  which  yc  is  the  distance  of  the  center  of  pressure  from  line  of 
intersection  of  the  plane  of  the  area  with  the  surface  of  the  liquid, 
y  is  the  distance  of  the  center  of  gravity  of  the  area  from  this 
line,  k  is  the  radius  of  gyration  of  the  area  with  respect  to  the 
same  line,  and  k0  is  the  radius  of  gyration  with  respect  to  a  paral- 
lel line  through  the  center  of  gravity.  These  equations  apply  to 
inclined  and  to  vertical  surfaces. 


CHAPTER  XVI 
MOTION 

156.  Displacement. — When  a  body  changes  its  position  relative 
to  another  body,  it  is  said  to  be  displaced  relatively  to  the  other 
body.     The  distance  from  the  initial  position  to  the  final  posi- 
tion is  the  displacement  of  the  body.     Displacement  is  generally 
measured  relative  to  the  earth.     Displacement  has  direction  as 
well  as  magnitude,  and  is,  therefore,  a  vector  quantity. 

Problems 

1.  A  man  steps  from  a  car  and  walks  east  a  distance  of  40  feet.     During 
the  same  time,  the  car  moves  west  60  feet.     What  is  the  displacement  of 
the  man  relative  to  the  earth  and  relative  to  the  car? 

Ans.     40  feet  east;  100  feet  east. 

2.  A  body  moves  from  one  position  to  another  position  which  is  40  feet 
east  and  30  feet  north  of  the  starting  point.     What  is  its  displacement? 

Ans.     50  feet  north  53£  08'  east. 

3.  A  base  ball  is  struck  by  a  bat.     Two  seconds  later  it  is  80^|^ajt  above  and 
150  feet  south  of  the  starting  point.     What  is  the  displacement? 

157.  Velocity. — Rate  of  displacement  is   called  velocity.     In 
problems  of  mechanics,  velocity  is  generally  expressed  in  feet 
per  second.     Since  displacement  is  a  vector,   and  velocity  is 
displacement  divided  by  time  (which  is  not  a  vector)  it  follows 
that  velocity  is  a  vector.     A  velocity  of  ten  feet  per  second  north 
is  not  equivalent  to  a  velocity  of  ten  feet  per  second  east.     When 
only  the  magnitude  of  the  motion  is  considered  without  reference 
to  its  direction,  the  rate  of  displacement  is  called  speed.     A  body 
moving  east  at  the  rate  of  ten  feet  per  second  has  the  same  speed 
as  a  body  moving  north  at  the  rate  of  ten  feet  per  second. 

Example 

A  man  is  40  feet  south  of  a  given  point  when  his  watch  reads  1  min.  20  sec. 
When  he  is  100  feet  south  of  the' point,  his  watch  reads  1  min.  35  sec. 
What  is  his  average  velocity  during  that  interval? 

The  displacement  is  100  —  40  =  60  feet  south.     The  time  interval  is  15 

f*f\ 

seconds.     Average  velocity  south  =  y^  =  4  feet  per  second. 

259 


260  MECHANICS  [ART.  157 

Velocity  is  calculated  by  dividing  the  difference  in  position 
by  the  corresponding  difference  in  time.  If  Xi  is  a  coordinate 
of  the  position  at  the  time  ti  and  Xz  is  a  coordinate  of  the 
position  at  the  time  £2,  and  if  vx  is  the  component  of  the  velocity 
parallell  to  the  X  axis, 

Vx  =  *L--L_?i.  Formula  XVII 

'2  —  t\ 

Similar  expressions  give  the  components  of  the  velocity  parallel 
to  the  other  axes  of  coordinates.  The  actual  velocity  is  the 
vector  sum  of  these  components. 

Formula  XVII  and  the  similar  expressions 
give  the  average  velocity  during  the  interval  of 
time.  The  actual  velocity  at  any  instant  may 
be  quite  different.  In  Fig.  234,  a  body  may 
start  at  A  and  move  along  the  curved  path 
ABCDE,  passing  over  each  interval  in  one 
second.  If  an  interval  of  one  second  is  con- 
sidered, the  velocity  during  the  first  second  is 
equal  to  the  length  A  B  and  is  in  the  direction 
from  A  to  B.  If  the  entire  4  seconds  are  con- 
sidered as  the  interval,  the  velocity  is,  apparently,  one-fourth  of 
the  length  A  E  and  is  in  the  direction  from  A  to  E.  Again,  the 
speed  or  magnitude  of  the  velocity  may  change.  A  body  may 
move  2  feet  in  one  second,  5  feet  in  the  next  second,  and  11  feet 
in  the  third  second.  For  the  first  second,  the  average  velocity  is 
2  feet  per  second;  for  the  first  two  seconds,  the  average  velocity 
is  3.5  feet  per  second;  for  the  three  seconds,  the  average  velocity 
is  6  feet  per  second.  When  the  velocity  varies  in  direction  and 
magnitude  and  it  is  desired  to  find  the  true  velocity  at  any 
instant,  a  small  displacement  and  a  correspondingly  small  inter- 
val of  time  must  be  employed.  Formula  XVII  becomes 


To  determine  experimentally  the  actual  velocity  at  any  given 
instant,  means  must  be  devised  for  measuring  small  intervals 
of  time  and  corresponding  increments  of  displacement.  Suppose 
it  is  desired  to  find  the  velocity  of  a  steel  ball  attached  to  a  small 
parachute  after  falling  two  seconds  from  rest.  The  ball  may 
fall  beside  a  vertical  scale  as  in  Fig.  235.  The  time  and  dis- 
placement may  be  measured  by  means  of  a  photographic  camera 


CHAP.  XVI]  MOTION  261 

provided  with  an  "  instantaneous  shutter "  of  known  time.  An 
electromagnetic  arrangement  may  be  provided  which  will  release 
the  shutter  two  seconds  after  the  ball  starts  to  fall.  If  the  ball 
is  properly  illuminated,  its  displacement  is  given  by  a  vertical 
streak,  AB  of  Fig.  235.  Suppose  that  the  "time"  of  the  shutter 
is  0.02  second,  and  that  the  length  of  the  streak  on  the  plate  is 
equivalent  to  0.36  foot  of  the  scale.  The  average 
velocity  for  the  0.02  second  is 


£.0 
.1 

—  .2 
— -      .3 

.4 
.5 

—  .6 

—  .7 

—  .8 

—  .9 

—  3.0 


0.36       10, 

v  =  ^-^r  =  18  feet  per  second.  • 
0-02 

This  experiment  gives  the  average  velocity  for 
the    interval    of    0.02   second.     If   the  change  of 
velocity   is    all    in   one   direction,   that   is,   if  the 
velocity  is  either  increasing  or  decreasing  through- 
out the  entire  interval,  the  average  velocity  is  the 
velocity  at  about  the  middle  of  the  interval.     The      pIG.  235. 
result  of  this  experiment  is,  then,  the  velocity  at 
about   2.01   seconds   after  the   beginning    of    the  fall.     If  the 
velocity  is  desired  at  2  seconds  from  the  beginning  of  the  fall, 
the   apparatus   must   be  arranged  to  release  the  shutter  1.99 
seconds  after  the  body  starts  to  move. 

Problems 

1.  The  position  of  a  body  is  given  by  the  equation  x  =  8£2,  in  which  x 
is  the  position  in  feet  and  t  is  the  time  in  seconds.     Find  the  approximate 
velocity  at  the  end  of  4  seconds  by  means  of  the  displacement  during  a  2- 

,  .   ,                                                      8  X  52  -  8  X  32       RAt. 
second  interval.         Ans.     Average  v  = ^ =  64  ft.  per  sec. 

2.  Solve  Problem  1  for  the  velocity  at  the  end  of  4  seconds  by  means  of 
the  average  velocity  during  an  interval  of  0.2  second. 

Ans.     Average  v  =  8  X  ^'^  ^  X  3'9*  =  64  ft.  per  sec. 

3.  The  position  of  a  body  is  given  by  x  =  2t*.     Using  an  interval  of  2 
seconds,  find  the  approximate  velocity  at  the  end  of  4  seconds. 

o   V   ^3   9    V   ^3 

Ans.     Average  v  =  =-^-  —~        -  =  98  ft.  per  sec. 

4.  Solve  Problem  3  for  an  interval  of  0.2  second. 

0(4.  1  3    Q  Q31) 

Ans.     Average  v  =  02  =  96'02  ft*  Per  sec" 

In  Problems  1  and  2,  the  average  velocity  for  the  2-second  interval  is 
the  same  as  the  average  velocity  for  the  0.2-second  interval.  It  is  fair  to 
assume  that  the  average  velocity  is  the  actual  velocity  at  the  middle  of  the 
interval  and  that  64  feet  per  second  is  the  true  velocity  at  the  end  of  4 
seconds. 


262  MECHANICS  [ART.  158 

In  Problems  3  and  4,  the  average  velocity  for  the  2-second  interval  is 
greater  than  the  average  velocity  for  the  0.  2-second  interval.  It  is  evident, 
therefore,  that  the  average  velocity  is  not  exactly  the  velocity  at  the  middle 
of  the  interval  and  that  a  very  short  interval  must  be  used  to  get  an  exact 
result. 

Problem 

6.  In  Problem  3,  let  the  interval  be  0.02  second  and  find  the  approximate 
velocity  at  the  end  of  4  seconds. 

Ans.     v  =  2(4'01Q  Q23'"  '*  =  ^6.0002  ft.  per  sec. 

It  is  evident  that  an  interval  of  0.2  second  is  sufficiently  short  to  give  a 
result  as  accurate  as  can  be  measured. 

168.  Average  and  Actual  Velocity.  —  Formula  XVII  gives  the 
average  velocity  for  the  interval.  From  the  problems  of  the 
preceding  article,  it  is  evident  that  the  average  velocity  is  some- 
times equal  to  the  velocity  at  the  middle  of  the  time  interval, 
and  at  other  times  is  not  equal  to  that  velocity.  A  few  cases  will 
now  be  considered  in  order  to  determine  under  what  conditions 
it  is  necessary  to  make  the  time  interval  very  short. 

Let  x  =  kt2  be  the  expression  for  the  displacement,  and  let  the 
average  velocity  be  found  for  an  interval  of  2AZ  seconds. 


Average  v  -  k(t  +  ^  ~  k(t  ~  ^  -  -  2kt 

2M  2&t 

The  result  is  independent  of  the  length  of  the  interval  2A£. 
Consequently,  when  the  displacement  varies  as  the  square  of  the 
time,  the  velocity  at  the  middle  of  the  interval  of  time  is  equal 
to  the  average  velocity  during  that  interval.  This  is  true  also 
when  the  displacement  varies  as  the  time. 

When  the  displacement  varies  as  the  cube  of  the  time,  x  =  kt3. 
For  an  interval  2At, 


average  t  .  _ 

It  is  evident  that  the  average  velocity  is  slightly  different 
from  the  velocity  at  the  middle  of  the  interval.  The  error, 
however,  decreases  rapidly  with  decrease  of  the  interval  of  time. 

In  an  experimental  problem,  it  may  be  known  that  the  dis- 
placement varies  as  the  square  of  the  time,  as  the  first  power  of 
the  time,  or  as  a  combination  of  the  two.  In  these  cases,  the 
time  interval  may  be  taken  of  any  convenient  length.  The 
average  velocity  during  the  interval  is  the  velocity  at  the  middle 


CHAP.   XVI]  MOTION  263 

of  the  interval.  If  the  experimental  values  of  the  velocity  are 
the  same,  whether  the  intervals  are  long  or  short,  it  may  be 
taken  for  granted  that  the  displacement  varies  as  the  time  or  as 
the  square  of  the  time. 

For  any  form  of  mathematical  expression  for  the  displacement 
in  terms  of  the  time,  it  is  easy  to  calculate  the  effect  of  the 
length  of  the  interval  upon  the  accuracy  of  the  result. 

159.  Velocity  as  a  Derivative.  —  When  the  interval  of  time 
becomes  infinitesimal,  Formula  XVII  reduces  to 

dx 

v*  =  dt' 
In  Problem  1  of  Art.  157, 


When  t  =  4  seconds,  vx  =  16  X  4  =  64  ft.  per  sec. 
In  Problem  3  of  Art.  157, 

x  =  2t 


When  t  =  4  seconds,  vx  =  6  X  16  =  96  ft.  per  sec. 

Since  these  derivatives  give  the  correct  result  so  easily,  it 
may  be  asked  why  the  methods  of  Article  157  are  employed. 
The  derivative  is  sufficient  when  the  equation  of  the  displace- 
ment in  terms  of  the  time  is  accurately  known.  In  most  cases, 
however,  the  displacement  is  not  known  in  terms  of  the  time 
and  an  experimental  determination  is  required  to  get  the  velocity. 
The  work  of  the  two  preceding  articles  is  intended  to  show  with 
what  accuracy  the  average  velocity  during  an  interval  gives  the 
true  velocity  at  the  middle  of  that  interval. 

Problems 

1.  The  position  of  a  body  is  given  by  the  equations  x  =  3t2,  y  =  t3,  in 
which  the  distances  are  expressed  in  feet  and  the  time  is  given  in  seconds. 
Find  the  magnitude  and  direction  of  the  resultant  velocity  when  t  =  4 
seconds.  Ans.     v  =  53.66  ft.  per  sec.  at  63°  26'  with  the  X  axis,  v 

2.  The  position  of  a  body  is  given  by  x  =  sin  t,  in  which  x  is  given  in 
feet  and  t  is  given  in  seconds.     Find  the  velocity  when  t  =  1  second  and 
when  t  =  2  seconds. 

Ans,     vi  =  0.540  ft.  per  sec.;  t>2  =  -  0.416  ft.  per  sec. 


264 


MECHANICS 


[ART.    159 


3.  The  position  of  a  body  is  given  by  x  —  r  cos  ut,  y  =  r  sin  ut,  in  which 
r  is  a  constant  length  expressed  in  feet  and  w  is  a  constant  number.  Find  the 
magnitude  and  direction  of  the  resultant  velocity  for  any  given  value  of  the 
time  /  / 

Ans.  vx  =  —  rw  sin  w£;  vy  =  rco  cos  co£;  v  =  rw,  at  an  angle  with  the  X 
axis  whose  tangent  is  equal  to  -cotan  at. 

Problem  3  gives  the  motion  of  a  body  moving  in  a  circle  of 
radius  r  with  uniform  speed  rco.  In  Fig.  236,  I,  the  body  is  at 
B  and  is  moving  in  a  counter-clockwise  direction.  The  radius 
from  the  center  to  B  makes  an  angle  cot  with  the  horizontal  line 
OA .  Fig.  236,  II,  is  the  vector  velocity  triangle.  The  horizontal 
component  vx  is  toward  the  left,  while  the  vertical  component  is 


y=  r  sin  cot  \ 


xH 

II 


FIG.  236. 


upward.  The  resultant  velocity  v  makes  an  angle  B  with  the 
horizontal  toward  the  right. 

Tan  B  =  —  cotan  coZ, 

which  shows  that  the  resultant  velocity  is  normal  to  the  line  OB. 
The  line  which  is  tangent  at  B  in  Fig.  236,  I,  gives  the  direction 
of  the  resultant  velocity. 

160.  Acceleration. — The  rate  of  change  of  velocity  is  called 
the  acceleration.  If  a  body  is  moving  east  with  a  velocity  of  10 
feet  per  second,  and  2  seconds  later  is  moving  east  with  a  velocity 
of  40  feet  per  second,  the  change  in  velocity  is  30  feet  per  second 
east.  This  change  takes  place  in  2  seconds.  The  rate  of  change 
is  30  -r-  2  =  15  feet  per  second  per  second. 


CHAP.  XVI]  MOTION  265 

Problems 

1.  If  the  acceleration  is  4  feet  per  second  per  second  east  and  the  body 
has  an  initial  velocity  of  12  feet  per  second  east,  what  will  be  its  velocity  5 
seconds  later? 

Ans.    v  =  32  ft.  per  sec.  east.  * 

2.  If  the  acceleration  is  4  feet  per  second  per  second  east  and  the  body 
has  an  initial  velocity  of  12  feet  per  second  west,  what  will  be  its  velocity 
4  seconds  later?  Ans.     v  =  8  ft.  per  sec.  east.  ^ 

3.  A  body  is  moving  north  20  feet  per  second.     Two  minutes  later  it  is 
moving  north  92  feet  per  second.     Find  its  acceleration.  > 

Ans.     36  ft.  per  sec.  per  min.;  0.6  ft.  per  sec.  per  sec.  * 

The  expression  for  the  component  of  the  acceleration  in  the 
direction  of  the  X  axis  is 

ax  =   v*  ~  v\  Formula  XVIII 

tz    —    t\ 

in  which  v\  is  the  component  of  the  velocity  in  the  direction  of  the 
X  axis  at  the  time  t\,  vz  is  the  component  at  the  time  Z2,  and  ax  is 
the  acceleration. 

Problems 

4.  A  car  is  moving  on  a  track  with  increasing  velocity.     By  means  of  a 
electro-magnet  actuated  by  a  current  through  a  break-circuit  chronometer, 
a  mark  is  made  on  the  track  every  2  seconds.     At  0  seconds  the  mark  is 
16  feet  from  a  reference  point.     At  2  seconds  the  mark  is  46  feet  from  the 
reference  point,  and  at  4  seconds  the  mark  is  100  feet  from  the  reference 
point.     Find  the  acceleration. 

The  average  velocity  for  the  first  2-second  interval  is  15  feet  per  second. 
For  the  next  interval,  the  average  velocity  is  27  feet  per  second.  Assuming 
that  the  average  velocity  is  the  velocity  at  the  middle  of  the  interval,  the 
the  velocity  at  1  second  is  15  feet  per  second  and  the  velocity  at  3  seconds  is 
27  feet  per  second.  The  acceleration  is 

a  =  -       -=—  =  6  ft.  per  sec.  per  sec. 
o  —  1 

The  problem  may  be  arranged  in  this  way: 

Time  Displacement         Velocity  Acceleration 

0  16 

1 15 

2  46  6  ft.  per  sec.  per  sec. 

3 27 

4  100 

6.  In  an  experiment  similar  to  Problem  4,  the  readings  were  1  second,  24 
feet;  3  seconds,  60  feet;  5  seconds,  140  feet.  Find  the  acceleration. 

6.  In  an  experiment  similar  to  Problem  4,  the  readings  were  1  second, 
20  feet;  3  seconds,  54  feet;  7  seconds,  218  feet.  Find  the  acceleration. 

Ans.     8  ft.  per  sec.  per  sec. 


266  MECHANICS  [ART.   161 

7.  In  Problem  4,  suppose  that  the  acceleration  is  constant.     Find  the 
position  at  the  6th  second. 

8.  In   Problem   1  of  Art.    157,  find  the  velocity  at  4  seconds  and  at 
5  seconds  by  means  of  intervals  of  0.2  second;  then  find  the  average  accel- 
eration for  the  one-second  interval. 

Ans.     Average  a  =  16  ft.  per  sec.  per  sec. 

9.  In  problem  3  of  Art.  157,  find  the  approximate  velocity  at  4  seconds 
and  at  6  seconds  by  means  of  2-second  intervals,  and  then  find  the  average 
acceleration  for  the  2-second  interval. 

Ans.     a  =  60  ft.  per  sec.  per  sec. 

10.  A  cannon  ball  acquires  a  velocity  of  1800  feet  per  second  in  0.004  second. 
Find    its    acceleration.  Ans.     a  =  450,000  ft.  per  sec.  per  sec. 

11.  A  baseball  with  a  speed  of  100  feet  per  second  is  stopped  in  0.001 
second.     Find  its   acceleration. 

12.  The  velocity  of  a  body  is  changed  from  80  feet  per  second  east  to  40 
feet  per  second  west  in  3  seconds.     Find  its  acceleration. 

Ans.     a  =  40  ft.  per  sec   per  sec.  west. 

161.  Acceleration  as  a  Derivative. — When  the  increment  of 
time  and  the  corresponding  increment  of  velocity  are  very  small, 
Formula  XVIII  becomes 

—  t-  w 

Since  vx  =  -jr*  Equation  (1)  reduces  to 


The  acceleration  along  the  other  coordinate  axes  is  given  by 
similar  derivatives. 

dvx  _  d2x 
"  ~dt  "".  ~dt2' 

av  =  -~~  =  -^;  Formulas  XIX 

_  dv,  _  d2z 
"   dt""   dt2 

In  some  works  of  mathematical  physics  the  components  of  velocity  and 
acceleration  are  written  thus:  The  component  of  the  velocity  parallel  to 
the  X  axis  is  x  and  the  component  of  the  acceleration  is  x.  A  derivative 
with  respect  to  time  is  written  with  a  single  dot  over  the  variable.'  The 
second  derivative  with  respect  to  time  is  written  two  dots  over  the  variable. 

Problems 

1.  The  position  of  a  body  is  given  by  x  =  ktz.  When  the  displacement 
varies  as  the  square  of  the  time,  show  that  the  acceleration  is  constant. 

Ans.     a  =  2k. 


CHAP.   XVI]  MOTION  267 

2.  Find  the  acceleration  in  Problem  1  of  Art.  157  by  means  of  the  second 
derivative.  Ans.     a  =  16  ft.  per  sec.  per  sec. 

3.  Find  the  acceleration  in  Problem  3  of  Art.  157  when  t  =  5  seconds. 

Ans.     a  =  12t  =  60  ft.  per  sec.  per  sec. 

The  answer  of  Problem  3  is  the  same  as  the  answer  of  Problem  9  of  the 
preceding  article  for  the  average  acceleration  for  the  interval  from  4  seconds 
to  6  seconds.  While  the  average  velocity  differs  considerably  from  the 
velocity  at  the  middle  of  the  interval,  the  acceleration  obtained  is  the 
correct  value  for  the  middle  of  its  assumed  interval. 

4.  The  position  of  a  body  is  given  by  the   equation  y  =  e*.     Find  the 
velocity  and  acceleration  when  t  —  2  seconds. 

Ans.     vy  =  ez  =  7.39;  ay  =  7.39. 

6.  The  position  of  a  body  is  given  by  the  equation  x  =  10  sin  4t.  Find 
the  velocity  and  acceleration  when  t  =  1  second  and  when  t  =  2  seconds. 

162.  Acceleration  as  a  Vector. — Since  velocity  is  a  vector  and 
acceleration  is  velocity  divided  by  time  (which  is  not  a  vector), 
it  follows  that  acceleration  is  a  vector  quantity.  Accelerations 
may  be  resolved  into  components  or  combined  into  resultants 
in  the  same  way  as  forces  or  other  vectors. 


Problems 

\/ 

1.  The  position  of  a  body  is  given  by  the  equations  x  =  2t  +  3t2,  y  =  t3. 
Find  the  coordinates  of  its  position  and  the  direction  and  magnitude  of 
its  velocity  and  acceleration  when  t  =  4  seconds.     Plot  the  curve  of  the 
path  of  the  body,  the  curve  of  its  velocity,  and  the  curve  of  its  acceleration. 

2.  The  position  of  a  body  is  given  by  x  =  4  cos  t,  y  =  4  sin  t.     Find  the 

components  of  the  velocity  and  acceleration  when  t  =  0,  when  t  =  ~  and 

when  t  =  IT. 

tf  ••.••     \  V  y  J 

Ans.     t  =  0,  vx  =  0;  vy  =  4;  ax  =  —4;  av  =  0. 

t  =  7j»  vx  =  —4;  vu  =  0;  ax  =  0;  ay  =  —4. 
t  =  TT,  vx  =  0;  %  =  —4;  ax  =  4;  au  =  0. 

Problem  2  is  that  of  a  body  which  moves  in  the  circumference 
of  a  circle  of  4-foot  radius  with  a  speed  of  4  feet  per  second. 
When  t  =  0,  the  body  is  at  position  A  of  Fig.  237.  The  motion 
is  directly  upward.  The  vertical  component  of  the  velocity  is 
the  entire  velocity  of  4  feet  per  second,  while  the  horizontal 
component  is  zero.  Since  the  vertical  component  is  not  chang- 
ing, ay  =  0.  The  direction  of  the  velocity  as  a  vector  is  continu- 
ally changing.  When  the  body  is  at  a  small  distance  above  A, 


268 


MECHANICS 


[ART.   162 


its  velocity  has  a  component  toward  the  left.     There  is  an  ac- 
celeration toward  the  left  to  give  it  this  change  of  velocity.     This 

acceleration  is  4  ft.  per  sec.  per  sec.     At  B,  where  t  =  ^  the 

entire  velocity  is  horizontal 
toward  the  left.  It  is  rep- 
resented by  vx  =  —  4,  v  y=  0. 
The  acceleration  is  4  ft.  per 
sec.  per  sec.  downward. 

The  results  of  Problem  2 
show  that  the  acceleration  is 
constant  in  magnitude  and  is 
directed  toward  -the  center, 
while  the  velocity  is  constant 
in  magnitude  and  is  directed 
along  the  tangent. 


FIG.  237. 


Problem 


3.  The  position  of  a  body  is  given  by  the  equations,  x  =  r  cos  cof ,  y  =  r  sin  co  t, 
in  which  r  and  co  are  constants.  Find  the  magnitude  and  direction  of  the 
acceleration  for  any  given  time. 


FIG.  238. 

Ans.     ax  =  —  rco2  cos  o)t;  ay    =  —  rco2  sin  (at;  resultant  acceleration  =  rco2. 
Problem  3  gives  the  motion  of  a  body  which  moves  in  a  circle  of  radius 
r  with  a  linear  velocity  of  r  co.     (Problem  3,  Art.  159).     If  the  linear  velocity 
rco  =  v, 

r2co2       vz 
acceleration  =  rco2  =  =  —  •  Formula  XX 


CHAP.   XVI]  MOTION  269 

Figure  238,  II,  is  the  vector  triangle  of  acceleration  for  Problem 
3  when  the  body  is  at  position  B  of  Fig.  238,  I.  The  horizontal 
component  is  negative,  and  is,  therefore,  directed  toward  the 
left.  The  vertical  component  is  negative,  and  is,  therefore, 
directed  downward.  If  0  is  the  angle  which  the  resultant 
acceleration  makes  with  the  horizontal, 

—  rco2  sin  ut 

tan  4>  =  ~     — »—    —.  =  tan  ut. 

—  rw2  cos  ut 

When  the  body  is  at  B  of  Fig.  238,  I,  the  acceleration  is  in  the 
direction  BO.  When  a  body  is  moving  in  a  circle  with  constant 
speed,  its  acceleration  is  constant  in  magnitude  and  is  directed 
toward  the  center  of  the  circle.  The  acceleration  is  due  to  the 
change  of  direction  of  the  velocity  and  not  to  change  of  magnitude. 
If  a  body  is  moving  in  any  curved  path  with  uniform  speed, 
the  acceleration  is  directed  toward  the  center  of  curvature  of  the 

v2 
path.     The  magnitude  of  the  acceleration  is  — ;  in  which  r  is  the 

radius  of  curvature  of  the  path  at  the  given  position. 

4.  A    body    is    moving    east    20   feet   per 
second.     Two    seconds    later    it    is   moving 
north  20  feet  per  second.     Find  the  average 
acceleration  during  the  2-second  interval. 

Figure  239  is  the  vector  diagram  of  ve- 
locities. The  increment  of  velocity  is  Av 
which  must  be  added  to  Vi  to  get  v2.  The 
vector  equation  is 

Vi  4-  Av  =  vj;  v2  - 
By  trigonometry, 

Av  =  28.28  ft.  per  sec.  northwest. 

28  28 
a  =  — ^ —  =  14.14  ft.  per  sec.  per  sec.  directed  northwest. 

5.  A  base  ball  is  pitched  horizontally  with  a  velocity  of  100  feet  per  second. 
It  leaves  the  bat  with  a  velocity  of  120  feet  per  second  at  an  angle  of  30  de- 
grees with  the  horizontal  and  passes  directly  over  the  pitchers  box.     What 
is  the  change  of  velocity?     If  the  ball  is  in  contact  with  the  bat  for  0.02 
second,  what  is  the  acceleration? 

Ans.     a  =  10,628  ft.  per  sec.  per  sec.  at  16°  24'  with  the  horizontal. 

6.  A  ball  is  thrown  against  a  wall  with  a  velocity  of  100  feet  per  second 
and  rebounds  in  the  opposite  direction  with  a  velocity  of  80  feet  per  second, 
what  is  the  acceleration? 

7.  Solve  Problem  5  if  the  ball  passes  directly  over  first  base. 

8.  A  body  moves  in  the  circumference  of  a  circle  of  radius  r  feet  per  second 
with  a  velocity  of  v  feet  per  second.     Find  the  direction  and  magnitude  of 
the  acceleration  as  a  vector. 


270 


MECHANICS 


[ART.   163 


Figure  240,  I,  shows  the  circle  of  Problem  8.  When  the  body 
is  at  A,  the  velocity  is  Vij  when  the  body  is  at  B,  the  velocity  is 
v2.  The  velocities  Vi  and  v2  have  equal  magnitudes.  Their 
directions,  on  the  other  hand,  are  different.  Since  Vi  is  normal  to 
the  radius  OA  and  v2  is  normal  to  the  radius  OB,  the  angle 

between  Vi  and  V2  is  equal  to 
the  angle  AOB.  Figure  240, 
II,  is  the  vector  diagram  of  the 
velocities.  The  change  in  ve- 
locity is  the  vector  Aw.  If 
this  change  takes  place  in  time 

A£,  the  acceleration  is  ^j    The 

direction  of  the  acceleration  is 
the  direction  of  Aw.  From  Fig. 
240,  II,  Aw  =  wA0,  in  which  v  is 
the  magnitude  of  the  velocities 
Vi  and  v2.  The  distance  from  A  to  B  of  Fig.  240,  I,  measured 
along  the  arc  of  the  circle,  is  rA0.  The  body  moving  with  a 
speed  v  passes  over  this  distance  in  time  A£. 


FIG.  240. 


/*A0  =  vAtj   At  = 

Aw 
a  =  — -  =  wA0 

At  w 


rA0 

v    ' 

rA0  =  w2 
r 


Formula  XX 


The  direction  of  Aw  is  perpendicular  to  the  bisector  of  the  angle 
between  YI  and  v2  and  is,  therefore,  parallel  to  the  bisector  of  the 
angle  AOB.  As  A0  is  made  smaller  and  smaller,  the  direction 
of  the  acceleration  is  always  parallel  to  this  bisector.  When  the 
angle  becomes  infinitesimal,  it  is  evident  that  the  acceleration  is 
directed  toward  the  center  of  curvature.  The  methods  of  Prob- 
lem 3  differ  from  those  of  Problem  8.  The  results,  however,  are 
identical. 

163.  Dimensional  Equations. — A  dimensional  equation  is  an 
equation  which  shows  with  what  powers  the  fundamental  quanti- 
ties of  mass,  length,  and  time  appear  in  the  expression  for  a  physi- 
cal quantity.  In  these  equations,  mass  is  represented  by  M, 
time  is  represented  by  T,  and  length  is  represented  by  L.  Some 
dimensional  equations  are: 

Area  =  L2; 
Volume  =  L3; 


CHAP.  XVI]  MOTION  271 

Density  =  ^  =  ML~*-, 
Velocity  =  ^  =  LT~l; 
Acceleration  =  -=-2  =  LT~2. 

Negative  exponents  are  generally  used  instead  of  fractional  forms. 
Dimensional  equations  have  two  important  uses.     These  are: 

1.  A  dimensional  equation  may  be  used  to  determine  whether 
a  literal  expression  for  a  quantity  is  possibly  correct. 

2.  In  transferring  from  one  system  of   units  to  another,  a 
dimensional  equation  indicates  the  required  power  of  the  ratio 
of  the  units  of  the  two  systems. 

The  dimensions  of  an  area  are  A  =  L2.  If  an  area  in  square 
yards  is  transferred  to  square  feet,  the  ratio  of  the  number  of 
units  of  length  in  feet  to  the  number  of  units  in  yards  is  3. 
Since  L  in  the  dimensional  equation  is  squared,  it  follows  that  the 
number  of  square  feet  in  a  given  area  is  32  times  as  great  as  the 
number  of  square  yards.  This  is  an  example  of  the  second  use 
of  a  dimensional  equation. 

If  a,  6,  and  c  are  lengths,  an  area  may  be  expressed  by  a2,  62, 
c2,  ab,  ac,  be,  c\/a2  +  fr2,  etc.  If  an  expression  which  is  known  to 
be  an  area  has  any  of  these  forms,  or  any  other  form  which  is 

equivalent  to  the  square  of  a  length,  the  expression  represents 

a2  _j_  52       

a  area.     Such  expressions  as  abc,  -        — ,  v  abc,  etc,  can  not 

C 

represent  areas.  If  such  an  expression  appears  in  a  quantity 
which  is  known  to  be  an  area,  it  will  be  recognized  as  incorrect. 
These  are  examples  of  the  first  use  of  a  dimensional  equation. 

Only  quantities  having  the  same  dimensions  may  be  added  or 
subtracted.  Volume  can  not  be  added  to  area  nor  velocity  to 
density.  If  a,  b,  and  c  are  lengths,  and  m  is  a  mass,  ab  +  b2  + 
a\/bc  and  me  +  m-\/a2  +  c2  are  possible  additions. 

Problems 

1.  The  dimension  of  the  trigonometric  functions  is  zero,  since  each  func- 
tion is  the  ratio  of  two  lengths.  If  a,  b,  c,  and  s  are  lengths,  is  it  possible 

6  ab  ,         1(8  -&)(»-  c)  s2  +  6c0 

to  represent  a  sine  by  — .  .  by  — ,  by  \/ : ,  or  by ( 

-y/a2  _|_  C27        C2  \  far a 

What  is  the  nature  of  the  quantity  expressed  by  \/s(s  —  a)(s  —  b)(s  —  c)? 
What  is  the  quantity  expressed  by  *y ? 


272  MECHANICS  [ART.  164 

2.  If  w  represents  mass,  q  represents  time,  and  a,  6,  and  c  represent  lengths, 
state  whether  the  following  are  possible  physical  quantities,  and  give  the 
meaning  of  those  which  are  possible: 

v,_    Vab    a2  +  fr2    a2  +  b2  +  c2 
abc       q    '        w  q2 

3.  What  is  the  coefficient  required  to  reduce  area  in  square  feet  to  area 
in  square  inches  and  to  area  in  square  yards?  Ans.     122;  (£)2. 

4.  What  is  the  factor  required  to  reduce  pounds  per  square  inch  to  grams 
per  square  centimeter? 

5.  How  is  velocity  in  feet  per  second  reduced  to  velocity  in  miles  per  hour? 

6.  How  is  acceleration  in  feet  per  minute  per  minute  reduced  to  accelera- 
tion in  feet  per  second  per  second? 

164.  Summary.  —  Displacement  is  the  amount  of  change  of 
position.  Displacement  is  a  vector. 

The  displacement  per  unit  of  time  is  the  velocity.  Velocity 
is  a  vector.  The  component  of  the  velocity  parallel  to  the  X 
axis  is 

#2  —  #1  _  dx 
''   t2  -  <i  ==  dt' 

The  dimensional  equation  of  velocity  is  LT~l. 

Change  of  velocity  per  unit  of  time  is  acceleration.  Since 
velocity  is  a  vector,  the  change  of  velocity  may  involve  change 
in  direction,  change  in  magnitude,  or  change  in  both  direction 
and  magnitude.  Acceleration  is  a  vector.  The  component  of 
acceleration  parallel  to  the  X  axis  is 


When  a  body  is  moving  in  a  circle  of  radius  r  with  a  velocity 
of  v  units  per  unit  of  time,  the  acceleration  toward  the  center  due 
to  the  change  of  direction  is 


The  dimensional  equation  of  acceleration  is  LT~2. 

A   dimensional  equation  is  useful  in  transferring  from  one 
system  of  units  to  another  and  in  checking  literal  equations. 


CHAPTER  XVII 
FORCE  AND  MOTION 

165.  Force  and  Acceleration. — If  the  forces  which  act  on  a 
body  are  in  equilibrium,  the  body  either  remains  at  rest  or  moves 
in  a  straight  line  with  constant  speed.  There  is  no  change  of 
motion.  Newton's  First  Law  states:  A  body  at  rest  remains  at 
rest  and  a  body  in  motion  continues  to  move  uniformly  in  a 
straight  line  unless  constrained  by  some  external  force  to  change 
that  condition.  If  a  horizontal  sheet  of  ice  were  perfectly  fric- 
tionless,  a  body  sliding  on  it  would  continue  to  move  in  one 
direction  with  constant  speed  for  an  indefinite  distance.  The 
attraction  of  the  earth  downward  on  the  body  and  the  reaction 
of  the  ice  upward  are  in  equilibrium  and  do  not  affect  the  motion. 
If  there  is  some  friction  between  the  body  and  the  ice,  the  velocity 
gradually  decreases  until  the  body  finally  comes  to  rest.  The 
force  of  friction  opposite  the  direction  of  the  motion  produces  a 
negative  acceleration  in  the  body.  The  ice  sheet  may  be  so 
inclined  to  the  horizontal  that  the  component  of  the  weight  down 
the  incline  exactly  equals  the  friction.  If,  then,  the  body  is 
given  an  initial  velocity  down  the  plane,  it  will  continue  to  move 
in  that  direction  with  constant  speed.  In  that  case,  the  forces 
are  in  equilibrium  and  cause  no  change  in  motion.  If  the  slope 
of  the  plane  is  increased  so  that  the  component  of  the  weight 
down  the  plane  exceeds  the  force  of  friction  in  the  opposite  direc- 
tion, the  velocity  of  the  body  down  the  plane  will  continually 
increase.  The  effective  force  is  down  the  plane  in  the  direction 
of  the  initial  velocity  and  the  acceleration  is  positive. 

The  acceleration  of  a  body  is  proportional  to  the  resultant 
effective  force  which  acts  on  it  from  other  bodies  and  is  inversely 
proportional  to  its  mass.  If  an  unbalanced  force  of  1  pound 
accelerates  a  given  mass  5  feet  per  second  per  second,  a  force  of  2 
pounds  will  accelerate  the  mass  10  feet  per  second  per  second 
and  a  force  of  3  pounds  will  accelerate  it  15  feet  per  second  per 
second.  If  a  given  force  accelerates  a  mass  of  1  pound  16  feet 
per  second  per  second,  that  force  will  accelerate  a  mass  of  2 
is  273 


274  MECHANICS  [ART.  165 

pounds  8  feet  per  second  per  second  and  will  accelerate  a  mass  8 
pounds  2  feet  per  second  per  second.  These  statements  are 
based  on  experiments  and  on  deductions  from  observed  facts. 
The  acceleration  is  in  the  direction  of  the  resultant  force.  New- 
ton's Second  Law  states:  Change  of  motion  is  proportional  to  the 
force  applied  and  takes  place  in  the  direction  in  which  the  force 
acts.  Expressed  algebraically, 

a  constant  X  effective  force 


acceleration  = 


mass  of  the  body 
kP 


P  =  ~->  Formula  XXI 

/C 

in  which  P  is  the  unbalanced  force,  a  is  the  acceleration  in  the 
direction  of  the  force,  m  is  the  mass,  and  k  is  a  constant.  The 
value  of  the  constant  k  depends  upon  the  units  of  force,  mass, 
length,  and  time  which  are  employed.  These  units  may  be  so 
chosen  that  k  is  equal  to  unity. 

The  most  common  case  of  uniformly  accelerated  motion  is  that 
of  a  freely  falling  body.  If  the  density  of  the  body  is  sufficiently 
great  so  that  the  resistance  of  the  air  may  be  neglected  without 
appreciable  error  (or  if  the  body  falls  in  a  vacuum),  the  only 
force  acting  on  the  body  is  the  attraction  of  the  earth.  The 
effective  force  is  the  weight  of  the  body.  Experiments  show  that 
the  acceleration  of  a  freely  falling  body  is  a  little  over  32  feet 
per  second  per  second.  This  figure  is  called  the  acceleration  of 
gravity  and  is  represented  in  algebraic  equations  by  the  letter  g. 

If  a  freely  falling  body  has  a  mass  of  one  pound,  the  effective 
force  (which  is  equal  to  its  weight)  is  one  pound.  A  force  of  one 
pound  gives  to  a  mass  of  one  pound  an  acceleration  of  g  feet  per 
second  per  second.  If  the  mass  of  a  freely  falling  body  is  m,  its 
weight  is  m  pounds.  Substituting  in  Equation  (1) 

km 
a  =  g  =  —  =  k.  (2) 

When  the  effective  force  is  expressed  in  pounds,  the  mass 
is  expressed  in  pounds,  and  the  acceleration  is  given  in.  feet  per 
second  per  second,  the  constant  k  of  Equation  (1)  and  Formula 
XXI  is  equal  to  g.  In  general,  when  the  unit  of  force  is  taken 
as  the  weight  of  the  unit  of  mass,  as  is  done  in  practically  all 


CHAP.  XVII]  FORCE  OF  MOTION  275 

engineering  and  commercial  work,  the  constant  k  is  equal  to  g 
and  Formula  XXI  becomes 

P  =  —  Formula  XXII 

y 

If  the  acceleration  is  given  in  feet  per  second  per  second,  the 
" standard"  value  of  g  is  32.174.  Formula  XXII  becomes 
effective  force  in  pounds  = 

mass  in  pounds  X  accel.  in  ft.  per  sec.  per  sec. 
32.174 

The  standard  value  of  g  is  the  acceleration  of  a  freely  falling 
body  at  the  sea  level  at  45  degrees  latitude.  It  is  sometimes 
written  gQ.  The  average  g  for  the  British  Isles  is  about  32.2. 
This  value  is  largely  used  in  English  books.  The  average  g  for 
The  United  States  is  less  than  32.174.  The  value  of  32.16  is  in 
common  use. 

Formula  XXII  may  be  written 

P  =  ™ag  (3) 

This  form  of  the  equation  states  that  the  effective  force  is  equal 
to  the  product  of  the  mass  multiplied  by  the  ratio  of  its  accelera- 
tion to  the  acceleration  of  gravity.  If  a  10-pound  shot  is  accel- 
erated 16  feet  per  second  per  second,  the  effective  force  is  10 
X  f|  =  5  pounds. 

Examples 

Solve  these  examples  without  writing,  using  32  feet  per  second  per  second 
as  the  value  of  g. 

1.  A  mass  of  24  pounds  has  an  acceleration  of  8  feet  per  second  per 
second.     What  is  the  effective  force? 

2.  A  mass  of  48  pounds  moving  with  a  velocity  of  80  feet  per  second  is 
brought  to  rest  in  5  seconds.     What  is  the  acceleration  and  what  is  the 
effective  force? 

3.  A  force  of  20  pounds  acts  on  a  10-pound  mass  for  2  seconds.     What 
is  the  acceleration  and  what  velocity  will  the  body  acquire  if  initially  at  rest? 

4.  A  force  of  12  pounds  acting  for  3  seconds  gives  a  body  a  velocity  of 
24  feet  per  second.     The  body  was  at  rest  at  the  beginning  of  the  interval. 
What  is  its  mass  in  pounds? 

Problems 

Useg  =  32.174 

1.  A  mass  of  40  pounds  moving  at  the  rate  of  60  feet  per  second  on  a 
horizontal  plane  is  brought  to  rest  in  6  seconds  by  the  friction  of  the  plane. 
Find  the  retarding  force  in  pounds  and  the  coefficient  of  friction. 

Ans.    P  =  12.43  Ib. 


276  MECHANICS  [ART.   166 

2.  A  body  is  moving  on  a  horizontal  plane  with  a  velocity  of  50  feet  per 
second.     If  the  coefficient  of  friction  is  0.3,  what  is  the  negative  accelera- 
tion and  in  what  time  will  the  body  come  to  rest? 

a  =  9.65  ft.  per  sec.  per  sec. 

3.  A  body  is  placed  on  a  smooth  inclined  plane,  which  makes  an  angle 
of  20  degrees  with  the  horizontal.     What  is  the  effective  force  in  pounds, 
and  what  is  the  acceleration  down  the  plane? 

Ans.     P  =  0.342  m.;  a  =  11.00  ft.  per  sec.  per  sec. 

4.  A  body  weighing  30  pounds  is  placed  on  an  inclined  plane,  which  makes 
an  angle  of  25  degrees  with  the  horizontal.     The  coefficient  of  friction  is  0.12. 
Find  the  effective  force  down  the  plane  and  find  the  acceleration. 

Ans.     P  =  9.42  lb.;  a  =  10.10  ft.  per  sec.  per  sec. 

6.  The  projectile  from  a  12-inch  naval  gun  weighs  1000  pounds  and  ac- 
quires a  velocity  of  2800  feet  per  second  in  0.035  second.  If  the  pressure  is 
constant  throughout  this  interval,  find  the  total  pressure  and  the  pressure 
per  square  inch.  Ans.  Total  pressure  =  2,486,000  lb. 

166.  Constant  Force. — When  the  resultant  force  is  constant, 
the  acceleration  is  constant  and  the  velocity  and  displacement 
are  easily  calculated.  If  a  is  the  acceleration,  the  change  of 
velocity  in  t  seconds  is  at  feet  per  second.  If  v0  is  the  velocity 
at  the  beginning  of  the  time  t,  the  velocity  at  the  end  of  the 
interval  is 

v  =  v0  +  at.  (1) 

Unless  the  initial  velocity  and  the  acceleration  are  in  the  same 
direction,  Equation  (1)  must  be  treated  as  a  vector  equation. 
R  In  Fig.  241,  time  is  plotted  as  ab- 

scissas and  velocity  as  ordinates.  The 
velocity  curve  is  the  straight  line  AB 
with  slope  equal  to  the  acceleration. 

Displacement  is  the  product  of  ve- 
locity multiplied  by  time.  When  the 
velocity  varies,  the  displacement  is  the 
product  of  the  average  velocity  mul- 

-i-l : l-i-i     tiplied    by   the   time  interval.     When 

¥10*241  ^e  accelerati°n  ig  constant,  as  in  Fig. 

241,  the  average  velocity  is  the  ve- 
locity at  the  middle  of  the  time  interval,  or  one-half  the  sum 
of  the  initial  and  final  velocities. 

,     .,          v0  +  v       2v0  -{-at  ,   at 

Average  velocity  =  — - —  =  -  -  =  v0  -f-  — •         (2) 

Z  2  2 

The  last  form  of  Equation  (2)  states  that  the  average  velocity, 


at 

4 


CHAP.  XVII]  FORCE  OF  MOTION  277 

when  the  acceleration  is  constant,  is  the  initial  velocity  plus  one- 
half  of  the  increase  during  the  interval  t. 

Examples 

Solre  wiihoijA  writing 

1.  The  velocity  of  a  body  changes  from  20  feet  per  second  east  to  40  feet 
per  second  east  in  5  seconds.     What  is  the  average  velocity  and  what  is 
the  distance  traversed  in  5  seconds? 

Ans.  Average  velocity  =  30  ft.  per  sec.  ;  displacement  =  30  X  5  =  150 
ft, 

2.  A  body  moving  with  a  velocity  of  40  feet  per  second  is  brought  to 
rest  in  4  seconds.     How  far  does  it  go?  Ans.     20  X  4  =  80  ft. 

3.  A  body  moving  40  feet  per  second  is  brought  to  rest  in  100  feet.     What 
is  the  time  interval,  and  what  is  the  acceleration? 

The  average  velocity  is  20  feet  per  second.  The  time  is  100  -5-  20  =  5 
seconds.  The  change  in  velocity  is  40  feet  per  second  and  takes  place  in 

5  seconds.     The  acceleration  is  —  40  -5-  5  =  —  8  ft.  per  sec.  per  sec. 

4.  In  Problem  3,  the  mass  is  60  pounds.     What  is  the  retarding  force  in 
pounds? 

5.  A  body  has  its  velocity  changed  from  20  feet  per  second  east  to  50 
feet  per  second  east  while  moving  140  feet.     Find  its  acceleration. 

6.  A  body  has  its  velocity  changed  from  40  feet  per  second  east  to  20 
feet  per  second  west  in  6  seconds.     What  is  its  average  velocity  for  the 

6  seconds?     How  far  is  it  from  the  starting  point  at  the  end  of  the  6  seconds? 
How  far  does  it  actually  travel  in  the  6  seconds? 

The  displacement  is  the  product  of  the  average  velocity  multi- 
plied by  the  time.  If  s  represents  the  displacement  in  any 
direction, 

,   «£2  /0\ 

V°  +'  (  ^ 


The  first  term  of  the  second  member  of  Equation  (3)  gives  the 
displacement  which  results  from  the  initial  velocity.  The 
second  term  gives  the  additional  displacement  which  results  from 
the  acceleration,  or  the  displacement  which  would  exist  if  the 
body  had  no  initial  velocity. 

Equation  (3)  may  be  represented  by  the  area  of  the  diagram  of 
Fig.  241.  The  rectangle  DAFC  represents  v0t;  the  triangle 

at2 
ABF  represents  -^- 

If  the  initial  velocity  and  the  acceleration  are  not  in  the  same 
direction,  Equation  (3)  must  be  regarded  as  the  sum  of  two 
vectors. 

In  solving  problems,  it  is  recommended  that  the  student  make 


278  MECHANICS  [ART.  166 

use  of  the  average  velocity  as  in  the  preceding  examples  and 
employ  Equation  (3)  as  little  as  possible. 

Problems 

1.  A  mass  of  40  pounds  is  moving  east  with  a  velocity  of  50  feet  per 
second.     After  an  interval  of  4  seconds,  the  body  is  120  feet  farther  east. 
Find  the  final  velocity,  the  acceleration,  and  the  effective  force.     Solve  by 
means  of  average  velocity  and  the  definition  of  acceleration.     Check  by 
Equations  (1)  and  (3). 

2.  A  40-pound  mass  has  a  velocity  of  60  feet  per  second  east.     It  is  sub- 
jected to  a  force  of  10  pounds  north.     What  will  be  its  position  and  velocity 
at  the  end  of  3  seconds? 

Ans.     Displacement  =  183.6  ft.  north  78°  37'  east  of  the  initial  point; 
v  =  64.68  ft.  per  sec.  north  68°  05'  east. 

3.  A  body  is  moving  east  with  a  velocity  of  50  feet  per  second.     After  6 
seconds,  it  is  moving  west  with  a  velocity  of  10  feet  per  second.     How  far 
does  it  move  during  the  6  seconds,  and  how  far  is  it  from  the  initial  point 
at  the  end  of  6  seconds?  Draw  a  diagram  similar  to  Fig.  241  to  give  its 
velocity  and  position. 

4.  A  body  starts  up  a  20-degree  inclined  plane  with  a  velocity  of  60  feet 
per  second.     The  coefficient  of  friction  is  0.2.     What  is  the  negative  accelera- 
tion when  the  body  is  going  up  the  plane?     In  what  time  will  it  come  to 
rest  and  what  will  be  its  distance  from  the  initial  point?     What  is  the 
acceleration  when  the  body  is  coming  down  the  plane?     How  long  will  it 
take  to  come  down  and  what  will  be  the  final  velocity  at  the  bottom? 

Ans.     Distance  =  105.57  ft. 

5.  Solve  Problem  4  if  the  coefficient  of  friction  is  0.4. 

6.  A  20-pound  mass  moving  with  a  velocity  of   100  feet  per  second  is 
stopped  in  2  feet.     Find  the  acceleration  and  the  effective  force. 

Ans.     a  =  2500  ft.  per  sec.  per  sec. ;     P  =  777  Ib. 

7.  A  base  ball  weighing  5.  5  oz.  and  moving  with  a  speed  of  100  feet  per 
second  is  stopped  in  1  foot.     Find  the  average  force. 

Ans.    P  =  53.4  Ib. 

8.  Solve  Problem  7  if  the  ball  is  stopped  in  1  inch. 

9.  A  cannon  ball  weighing  1000  pounds  acquires  a  velocity  of  2800  feet 
per  second  while  moving  50  feet.     Find  the  acceleration  and  the  effective 
force  on  the  projectile.  Ans.     P  =  2,436,000  Ib. 

10.  The  projectile  of  Problem  9  strikes  a  steel  plate  and  is  stopped  in 
2  feet.     Find  the  force. 

11.  An  automobile  is  traveling  on  a  level  street  with  a  velocity  of  30  feet 
per  second.     The  coefficient  of  starting  friction  is  0.35.     Half  the  weight 
comes  on  the  rear  wheels.     In  what  time  can  the  car  be  brought  to  rest  and 
how  far  does  it  go,  if  the  brakes  are  set  so  that  the  wheels  do  not  skid? 

Ans.     t  =  5.33  sec.;  s  =  80ft. 

12.  Solve  Problem  11  if  the  coefficient  of  sliding  friction  is  0.25  and  the 
brakes  are  set  too  tight  so  that  the  wheels  skid. 

13.  Solve  Problem  11  if  the  speed  is  only  15  feet  per  second. 

14.  Solve  Problem  11  for  a  speed  of  30  miles  per  hour,  15  miles  per  hour, 
and  10  miles  per  hour. 


CHAP.  XVII]  FORCE  OF  MOTION  279 

16.  An  elevator  starts  from  rest  and  is  accelerated  upward  6  feet  per 
second  per  second  for  3  seconds.  It  then  moves  upward  with  constant 
speed  for  5  seconds.  For  the  last  26  feet  the  acceleration  is  downward. 
What  is  the  total  distance?  What  is  the  effective  force  on  a  man  weighing 
160  pounds  during  each  stage  of  the  ascent?  What  is  the  pressure  between 
his  shoes  and  the  floor?  If  the  man  stands  on  a  platform  scale,  what  is 
his  apparent  weight  during  each  stage? 

16.  What  is  the  direction  of  the  acceleration  of  an  elevator  at  the  bottom 
when  it  is  starting  up,  and  when  it  is  coming  to  a  stop  at  the  bottom?  What 
is  the  direction  of  acceleration  at  the  top  when  it  is  starting  down,  and  when 
it  is  coming  to  a  stop  at  the  top? 

167.  Integration  Methods. — Equations  (1)  and  (3)  of  the 
preceding  article  were  derived  by  algebraic  methods.  These  are 
the  better  methods  for  constant  acceleration,  as  they  give 
greater  prominence  to  the  physical  ideas  than  do  the  methods  of 
calculus.  With  variable  acceleration,  calculus  is  necessary.  In 
order  to  become  familiar  with  the  application  of  calculus,  it  is 
advisable  to  apply  it  to  constant  acceleration  and  check  the 
results  by  means  of  the  equations  of  the  preceding  article. 

—  =  a  (1) 

Multiplying  by  dt  and  integrating 

v  =  at  +  Ci,  (2) 

in  which  Ci  is  an  integration  constant.     If  v0  is  the  velocity 
when  t  =  0,  C\  =  VQ  and  Equation  (1)  becomes 

v  =  v0  +  at,  (3) 

which  is  Equation  (1)  of  the  preceding  article. 

dx 
Substituting  v  =  -77  in  Equation  (3) 

dx  /A^ 

dt=»°  +  <*>  <4> 

x  =  vQt  +  -=-  H-  Cz-  (5) 

If  space  is  measured  from  the  position  where  t  =  0,  then  Cz  =  0, 
and 

,     a^  fa\ 

x  =  v0t  +  -£-•  (Q) 

2 

If  space  is  measured  from  some  other  point  such  that  x  =  x0 
when  t  =  0,  then  Cz  —  XQ  and 

x  =  x,  +  tut  +  ~-  (7) 


280 


MECHANICS 


[ART.    168 


Problem 

The  acceleration  of  a  body  is  proportional  to  the  time  and  is  8  feet  per 
second  per  second  when  t  =  2  seconds.  The  velocity  is  10  feet  per  second  in 
the  direction  of  the  acceleration  when  t  =  1  second.  By  integration,  find  the 
equations  of  velocity  and  displacement  in  terms  of  the  time  and  calculate 
the  velocity  and  displacement  when  t  =  5  seconds. 

Ans.  v  —  58.0  ft.  per  second;  s  =  123.3  ft.  from  the  position  when 
t  =  0. 

168.  Connected  Bodies. — When  the  equilibrium  of  connected 
bodies  is  considered,  the  entire  system  is  first  taken  as  a  free 
body  for  the  determination  of  the  external  reactions.  The 
system  is  then  divided  into  parts,  and  each  portion  is  treated  as 
a  free  body  in  equilibrium  under  the  action  of  the  external 
forces  and  of  the  internal  forces  from  the  adjacent  portions 
of  the  system.  This  method  is  used  in  the  calculation  of  the 
forces  in  trusses  and  other  structures.  In  a  similar  manner, 
Formula  XXII  may  be  applied  to  an  entire  system  or  to  any  part 
of  the  system. 

Example 

Figure  242  shows  a  20-pound  mass  on  a  smooth  horizontal  plane.  A 
cord  attached  to  the  mass  runs  horizontally  over  a  smooth  pulley  and 

supports    a    12-pound 

201       lib.  mass.      Assuming    that 

the  cord  and  the  pulley 
are  weightless  and  that 
the  plane  and  pulley  are 
frictionless,  find  the  ac- 
celeration and  the  tension 
in  the  cord. 

The  two  masses  and 
the  connecting  cord  form 
a  system.  The  weight  of 
the  20-pound  mass  and 
the  reaction  of  the  plane 
are  in  equilibrium.  The 
unbalanced  effective 
force  is  the  pull  of  the 
earth  on  the  12-pound 
mass.  The  pulley  merely 
changes  the  direction  of 
the  cord.  Mechanically 

the  system  is  equivalent  to  Fig.  242,  II,  in  which  both  masses  are  on  the 
same  horizontal  plane  and  a  horizontal  force  of  12  pounds  is  applied  to 
the  12-pound  mass 

Considering  the  system  as  a  whole,  the  effective  force  is  12  pounds  and 


III 


Fid.  242. 


CHAP.  XVII]  FORCE  OF  MOTION  281 

the  mass  which  must  be  accelerated  is  32  pounds.     Substituting  in  Formula 
XXII, 

12  =  3f; 

12  X  32.174 
a  =  — —  =  12.065  it.  per  sec.  per  sec. 

oZ 

To  determine  the  tension  in  the  cord,  which  is  an  internal  force  between 
portions  of  the  system,  the  12-pound  mass  may  now  be  treated  as  a  free 
body.  The  forces  which  act  on  this  free  body  are  the  pull  of  the  earth 
downward  and  the  tension  of  the  cord  upward.  The  effective  force  is  12  —  T 
pounds,  Fig.  242,  III.  The  mass  which  is  accelerated  by  this  force  is 

12  pounds.     The  acceleration  is  now  known  to  be  12.045  or  -^~  feet  per 

oZ 

second  per  second. 

12  X^ 
10       ™        12  X  12.065  ^32         ,  ,  „ 

12  -  T  =        32.174       Or^^"  =  4'51b'; 
T  =    12  -  4.5  =  7.5  Ib.  ' 

As  a  check,  the  20-pound  mass  may  be  considered  as  the  free  body.     The 
effective  force  is  T  pounds  toward  the  right  and  the  mass  is  20  pounds. 

20  X^ 
2°X 


Problems 

1.  Solve  the  example  above  if  the  coefficient  of  friction  between  the  20- 
pound  mass  and  the  plane  is  0.1. 

Ans.     a  =  --   3~  2\  =  10.054  ft.  per  sec.  per  sec.;  T  =  8.25  Ib. 

2.  A  40-pound  mass  is  placed  on  a  plane  which  makes  an  angle  of  20 
degrees  with  the  horizontal.     The  mass  is  attached  to  a  cord  which  runs 
up  parallel  to  the  plane,  passes  over  a  smooth  pulley,  and  supports  a  30- 
pound  mass  on  the  free  end.     Find  the  acceleration  and  the  tension  on  the 
cord  if  the  plane  is  smooth.     Check. 


Ans.     P  =  16.32  Ib.;  a  =  =  7.501  ft.  per  sec.  per  sec.;  T  =  23.02  Ib. 

3.  Solve  Problem  2  if  the  coefficient  of  friction  between  the  plane  and 
the  40-pound  mass  is  0.1. 

Ans.     a  =  5.776  ft.  per  sec.  per  sec.;  T  =  24.62  Ib. 

4.  What  velocity  will  the  bodies  of  Problem  2  acquire  in  3  seconds  and 
what  distance  will  they  travel  in  3  seconds  if  they  start  from  rest? 

5.  A  mass  of  20  pounds  is  on  a  horizontal  plane.     It  is  attached  to  a 
cord  which  runs  horizontally  over  a  smooth  weightless  pulley  and  sup- 
ports a  mass  of  16  pounds.     The  system  starts  from  rest  and  travels  24 
feet  in  the  first  2  seconds.     Find  the  coefficient  of  friction  between  the  20- 
pound  mass  and  the  horizontal  plane.  Ans.    f  =  0.129. 


282 


MECHANICS 


[ART.    169 


6.  Figure  243  shows  a  cord  which  runs  over  a  pulley  and  supports  a 
mass  of  6  pounds  on  one  end  and  a  mass  of  4  pounds  on  the  other.  If 
the  pulley  is  weightless  and  frictionless  and  if  the  cord  is  weightless,  what 
velocity  will  the  system  acquire  in  4  seconds  after  starting  from  rest?  What 
distance  will  it  travel  in  the  4  seconds?  What  is  the  tension  in  the  cord? 
What  is  the  total  load  on  the  pulley.  If  this  load  is 
less  than  10  pounds  explain  the  discrepancy? 

Figure  243  represents  the  Atwood  machine, 
which  is  considerably  used  to  demonstrate  the 
laws  of  acceleration  and  to  determine  g  approxi- 
mately. The  machine  is  subject  to  errors  which 
result  from  the  friction  and  from  the  mass  of 
the  pulley.  These  errors  may  be  determined 
experimentally  and  corrections  made.  Error 
caused  by  the  varying  length  of  cord  on  the  two 
sides  may  be  avoided  by  having  the  cord  con- 
tinuous as  is  shown  in  the  figure.  The  total 
mass  of  the  cord  must  be  added  to  that  of  the 
two  suspended  bodies  togefcmof  Formula  XXII. 


4lb. 


2lb. 


4lkx' 


FIG.  243. 


Problems 


7.  A  20-pound  mass  is  placed  on  a  plane  which  makes  an  angle  of  20 
degrees  with  the  horizontal.     The  mass  is  attached  to  a  cord  which  runs 
up  the  plane,  passes  over  a  smooth  weightless  pulley,  and  supports  a  mass 
of  12  pounds  on  the  free  end.     Starting  from  rest,  the  system  moves  18 
feet  during  the  first  3  seconds.     Find  the  coefficient  of  fric- 
tion between  the  plane  and  the  20-pound  mass. 

Ans.     F  =  1.1721b.;/  =  0.063. 

8.  In   Fig.   244,  the   movable   pulley   and  the   mass   W 
together   weigh  160   pounds.     The  mass  P  is  100  pounds. 
Find  the  acceleration   of  each  body  and  the  tension  in  the 
rope  if  the  mass  of  the  fixed  pulley  and  the  rope  is  negligible 
and  there  is  no  friction.  Ans.     T  =  85.71  Ib. 

Suggestion:  In  this  problem  it  is  best  to  write  a  sep- 
arate equation  for  each  of  the  two  bodies  and  combine  these 
equations  to  solve  for  T  and  a. 

169.  Velocity  and  Displacement. — The  equations 
of  Art.  167  give  the  displacement  in  terms  of  the 
initial  velocity,  the  acceleration,  and  the  time.     It     FlG  344 
is  sometimes  desirable  to  express  the  displacement 
in  terms  of  the  initial  and  final  velocities  and  the  acceleration, 
and  to  eliminate  the  time. 

+  v 
~2~ 


s  = 


(1) 


CHAP.  XVII]  FORCE  OF  MOTION  283 

From  the  definition  of  acceleration, 

V   —   V0 


t 


a 

v2  - 

s  =  — — 


t;2  -  v02  =  2  as.  (4) 

When  the  initial  velocity  is  zero,  Equation  (4)  becomes 

v2  =  2as,  Formula  XXIII 

If  the  final  velocity  is  zero, 

vl  =  2as,  (5) 

in  which  a  is  negative. 

Problems 

1.  The  acceleration  of  a  given  body  is  8  feet  per  second  per  second. 
What  is  its  velocity  after  moving  36  feet  from  the  position  of  rest?     Solve 
by  Formula  XXIII.     Check  by  means  of  the  time. 

2.  The  velocity  of  a  body  changes  from  4  feet  per  second  to  12  feet  per 
second  while  it  travels  32  feet.     Find  the  acceleration.     Solve  by  one  of 
the  equations  above  and  check  by  means  of  the  average  velocity  and  the 
time. 

3.  The  acceleration  of  a  body  is  12  feet  per  second  per  second.     How 
far  must  it  travel  in  order  that  the  velocity  may  change  from  4  feet  per 
second  to  16  feet  per  second  in  the  same  direction? 

4.  The  answer  to  Problem  3  is  the  same  whether  the  initial  velocities  are 
in  opposite  directions  or  in  the  same  direction.     What  is  the  meaning  of  s 
when  the  velocities  are  in  opposite  directions? 

170.  Energy.  —  A  force  of  P  pounds  acts  on  a  mass  of  m  pounds 
which  is  moving  with  a  Velocity  v  feet  per  second  in  the  direction 
of  the  force.  During  a  time  interval  dt,  the  displacement  is 
ds  =  vdt  and  the  work  done  by  the  force  of  P  pounds  is 

dU  =  Pds  =  Pvdt.  (1) 

The  force  P  accelerates  the  body  of  mass  m.     The  value  of  the 
force  in  terms  of  the  mass  and  the  acceleration  is 

ma  _mdv  ,  . 

g    "  gdt' 

Substituting  in  Equation  (1)  and  integrating, 


JTT          n     J*  /Q\ 

dU  =  P  vdt  =  -  •;  (3) 

t      ...  ** 

U  =  ^-  Formula  XXIV 

If  the  body  starts  from  rest,  the  limits  of  Formula  XXIV  are 


284  MECHANICS  [ART.  170 

0  and  v.  The  formula,  therefore,  gives  the  entire  work  done 
in  increasing  the  velocity  from  zero  to  v.  It  is  not  necessary 
that  the  force  and  acceleration  should  be  constant  in  order  that 
these  equations  may  be  valid. 

Formula  XXIV  is  the  expression  for  the  kinetic  energy  of 
the  body.  This  energy  depends  upon  the  mass  and  the  velocity. 
It  is  immaterial  how  the  velocity  is  obtained.  Since  vz  is 
positive  whether  v  is  positive  or  negative,  the  kinetic  energy  is 
independent  of  the  direction  of  the  motion.  Kinetic  energy  is, 
therefore,  a  scalar  quantity. 

In  Formula  XXIV,  if  the  mass  is  given  in  pounds  and  the 
velocity  in  feet  per  second,  g  is  32.174  and  the  energy  is  given  in 
foot  pounds. 

Kinetic  energy  in 

mass  in  pounds  (v  in  feet  per  sec.)2 
footpounds   •  2  X  32.174 

Problems 

1.  A  car  weighing  2400  pounds  is  moving  with  a  velocity  of  20  feet  per 
second.     Find  its  kinetic  energy  in  foot-pounds.     If  the  car  is  brought  to 
rest  while  moving  80  feet,  what  is  the  average  retarding  force? 

Ans.     Kinetic  energy  =  14,919  foot-pounds;  force  =  186.5  Ib. 

2.  Solve  Problem  1  if  the  velocity  is  40  feet  per  second. 

3.  The  car  of  Problem  1  has  its  velocity  changed  from  40  feet  per  second 
to  20  feet  per  second  while  it  runs  100  feet.     Find  the  force  required. 

4.  What  is  the  kinetic  energy  of  a  cannon  ball  weighing  1000  pounds  and 
moving  with  a  velocity  of  2800  feet  per  second.     If  this  projectile  is  fired 
from  a  gun  which  is  50  feet  in  length,  what  is  the  average  force? 

5.  The  projectile  of  Problem  4  strikes  a  wall  and  is  stopped  in  10  feet. 
Find  the  average  pressure. 

Problems  of  accelerated  motion  (whether  positive  or  negative 
acceleration)  in  which  the  distance  and  change  in  velocity  are 
given  are  best  solved  by  equating  the  change  of  kinetic  energy 
with  the  work  done  on  the  body  when  the  acceleration  is  positive 
or  with  the  work  done  by  the  body  when  the  acceleration  is 
negative.  Problems  in  which  the  time  and  the  change  in  velocity 
are  given  are  best  solved  by  means  of  the  acceleration  and 
Formula  XXII. 

Example 

A  mass  of  40  pounds  has  its  velocity  changed  from  60  feet  per  second  to 
20  feet  per  second  while  it  goes  200  feet.  Find  the  force  required. 

Change  in  kinetic  energy  =  40  %®®~*™  =  1989  =  200  P;  P  =  9.95  Ib. 


CHAP.  XVII]  FORCE  OF  MOTION  285 

To  solve  this  example  by  means  of  the  acceleration, 

80  +  20 
average  v  =     —  -  --  =  40  ft.  per  sec.; 

200 
*  =  —  =osec.; 

20  -  60 
a  =     —  =  —          -  8  ft.  per  sec.  per  sec.; 


p  - 


It  is  evident  that  the  method  of  kinetic  energy  is  the  shorter 
for  a  problem  in  which  the  distance  is  given. 

171.  Potential  Energy.  —  In  Art.  121,  potential  energy  was  de- 
nned as  the  energy  of  position.  If  a  12-pound  mass  is  lifted  a 
vertical  distance  of  10  feet,  work  done  on  the  mass  is  120  foot- 
pounds and  the  mass  can  do  120  foot-pounds  of  work  as  it  returns 
to  its  original  position.  The  change  of  potential  energy  is 
120  foot-pounds. 

If  no  energy  is  lost  by  change  into  heat  or  by  work  on  other 
bodies,  the  sum  of  the  potential  and  kinetic  energy  of  a  body  or 
system  of  bodies  remains  constant.  This  statement,  which  is 
based  on  experiments,  is  called  the  Law  of  Conservation  of  Energy. 
Energy  may  be  transformed  from  one  kind  to  another  but  can 
not  be  destroyed.  Many  problems  of  mechanics  may  be  solved 
by  means  of  this  principle. 

Example 

Solve  Problem  8  of  Art.  168  by  means  of  the  energy  relations.  Assume 
that  the  mass  of  100  pounds  moves  downward  a  distance  of  s  feet.  The 
work  is  100  s  foot-pounds  and  the  potential  energy  of  the  system  is  reduced 
that  amount.  At  the  same  time  the  mass  of  160  pounds  is  raised  a  distance 

o 

of  -Q  feet  and  t-he  potential  of  the  system  is  increased  80  foot-pounds.     The 

total  change  of  potential  energy  is  20  s  foot-pounds.     Equating  the  change 
of  potential  energy  with  the  change  of  kinetic  energy, 


From  Formula  XXIII, 

v2  =  2as; 

a  =      =  4.595  ft.  per  sec.  per  sec. 


286  MECHANICS  [ART.  172 

Problems 

1.  A  body  slides  50  feet  down  a  smooth  inclined  plane  which  makes  an 
angle  of  25  degrees  with  the  horizontal.     What  is  its  velocity  at  the  end  of 

its  descent?  Ans.     21.13  m  =  ^-;  v  =  36.8  ft.  per  sec. 

2.  The  body  of  Problem  1  weighs  20  pounds  and  the  friction  is  equivalent 
to  a  force  of  2  pounds.     Find  the  final  velocity. 

Ans.     322.60  ft.-lb.  =  ^*;  v  =  32.2  ft.  per  sec. 

3.  Solve  Problem  1  if  the  coefficient  of  friction  is  0.2. 

4.  A  block  and  tackle  has  three  ropes  which  support  the  weight  (Fig.  192). 
The  movable  pulley  and  its  load  weigh  240  pounds.     A  mass  of  120  pounds 
is  hung  on  the  free  end  of  the  rope.     Find  the  velocity  of  the  free  end  of  the 
rope  when  the  load  is  lifted  20  feet.     Find  the  acceleration  and  the  tension 
on  the  rope. 

6.  A  mass  of  3  pounds  is  placed  on  one  end  of  the  cord  of  an  Atwood 
machine  and  a  mass  of  2  pounds  is  placed  on  the  other  end.  The  efficiency 
of  the  pulley  is  95  per  cent.  What  is  the  velocity  after  moving  10  feet? 
What  is  the  tension  in  the  cord? 


172.  Motion  Due  to  Gravity. — When  a  body  is  not  supported, 
its  weight  is  the  effective  force,  and  its  acceleration  downward  is 
equal  to  g.  If  displacement  and  velocity  downward  are  taken  as 
positive,  then  the  force  of  gravity  and  the  acceleration  are  posi- 
tive. In  the  equations  of  the  preceding  articles  the  acceleration 
a  becomes  g  and 

v  ='v0  +  gt,  (1) 

h  =  v0t  +  ^  (2) 

v*  -  vl  =  2gh,  (3) 

in  which  h  is  the  vertical  distance  and  is  positive  downward. 
When  the  initial  velocity  is  zero,  Equation  (3)  becomes 

v*  =  2gh  Formula    XXIII 

This  form  of  Formula  XXIII  is  important.  It  gives  the  velocity 
which  a  body  will  acquire  in  falling  from  a  given  height,  the 
height  to  which  a  body  will  rise  with  a  given  initial  velocity 
upward,  and  the  velocity  of  flow  of  a  liquid  from  an  orifice  under 
a  given  head. 

When  the  initial  velocity  is  upward,  it  is  often  desirable  to 
regard  velocity  and  displacement  upward  as  positive.  The 
acceleration  of  gravity  is  then  taken  as  a  negative  quantity. 


CHAP.   XVII]  FORCE  OF  MOTION  287 

Example 

A  body  is  thrown  upward  with  a  velocity  of  80  feet  per  second.     What 
will  be  its  velocity  when  it  is  20  feet  above  the  starting  point? 
Using  Equation  (3)  with  h  positive  and  g  negative, 

v2  =  6400  -  1286.96; 
v  =  71.5  feet  per  second. 

Problems 

1.  In  the  example  above  there  are  two  solutions  for  the  velocity.    Explain. 

2.  A  body  is  thrown  vertically  upward  with  a  velocity  of  100  feet  per 
second.     How  high  will  it  rise?     In  what  time  will  it  return  to  the  starting 
point?  Ans.     h  =  155.4  ft.;  time  of  ascent  and  return  =  6.216  sec. 

Solve  Problems  3,  4,  and  5  without  writing.  Use  g  =  32  and  employ  the 
average  value  of  the  velocity  when  convenient. 

3.  A  body  is  thrown  upward  with  a  velocity  of  80  feet  per  second.     How 
long  will  it  rise?     What  is  the  average  velocity  upward?     How  high  will 
it  go? 

4.  A  body  is  thrown  upward  with  a  velocity  of  80  feet  per  second.     What 
is  the  average  velocity  during  the  first  second?     How  far  does  it  go  during 
the  first  second? 

6.  A  body  is  thrown  upward  with  a  velocity  of  40  feet  per  second.  What 
is  its  average  velocity  during  the  first  second?  Where  will  it  be  at  the  end 
of  the  first  second?  What  is  its  average  velocity  during  the  first  2  seconds? 
Where  will  it  be  at  the  end  of  2  seconds?  How  far  will  it  travel  during  the 
first  2  seconds? 

6.  Derive  Formula  XXIII  by  equating  the  potential  and  kinetic  energy. 

7.  A  body  is  thrown  upward  with  a  velocity  of  120  feet  per  second.     At 
what  point  will  the  velocity  be  60  feet  per  second?     At  what  point  will 
the  velocity  be  150  feet  per  second? 

8.  In  what  time  will  a  body  fall  100  feet  if  it  has  an  initial  velocity  of  60 
feet  per  second  downward? 

8.  Derive  Equation  (3)  by  means  of  kinetic  and  potential  energy. 

173.  Projectiles. — A  body  which  is  thrown  into  the  air  and 
continues  its  motion  under  the  action  of  no  external  force  except 
its  weight  and  the  resistance  of  the  air  is  called  a  projectile.  For 
the  present,  consideration  will  be  given  to  comparatively  heavy 
projectiles  moving  at  relatively  low  velocities.  The  resistance 
of  the  air  may  be  neglected  for  such  projectiles  and  the  only  force 
is  that  due  to  gravity.  The  problems  of  the  preceding  article  are 
examples  of  projectiles  with  initial  velocity  vertical. 

The  only  effect  of  gravity  is  to  change  the  vertical  component 
of  the  velocity.  The  horizontal  component  remains  constant. 
The  velocity  at  any  instant  is  the  resultant  of  the  horizontal  and 
vertical  components. 


288 


MECHANICS 


[ART.    173 


Figure  245  shows  positions  of  a  body  A  which  has  been  thrown 
horizontally.  At  the  same  time,  a  second  body  B  has  been 
allowed  to  fall.  Body  A  moves  over  equal  horizontal  distances 
during  each  interval  of  time,  and  falls  vertically  through  the  same 
distances  as  a  body  which  drops  directly  downward.  The  first 
body  A  may  be  thrown  from  a  spring  gun.  The  second  body  B 
may  be  supported  by  an  electromagnet  the  circuit  of  which  is 
broken  when  the  projectile  leaves  the  gun.  If  their  paths  do 
not  intersect,  both  bodies  will  strike  the  horizontal  floor  at  exactly 
the  same  instant.  If  their  paths  intersect,  the  bodies  will  collide. 


~    C/T 


r 


b 


X 


\  C 

\ 

\ 
\ 
\ 
\ 
\ 

\4 


-<j>  B 


B 


• 

•g  B 


-Q  B 


FIG.  245. 


In  Fig.  ,246,  the  first  body  is  thrown  at  an  angle  with  the 
horizontal  and  the  second  body  is  dropped  at  the  same  time  from 
a  point  which  is  in  the  line  of  the  initial  velocity  of  the  first  body. 
The  distance  which  the  projectile  falls  from  the  line  of  its  initial 
velocity  is  exactly  the  same  as  in  the  case  when  the  initial  velocity 
is  horizontal  or  zero.  The  projectile  may  be  regarded  as  moving 
with  constant  speed  along  the  line  of  its  initial  velocity  and  falling 
from  that  line  with  constant  acceleration.  If  the  path  of  the 
projectile  and  the  falling  body  intersect,  they  will  collide.  If  the 
paths  do  not  intersect,  they  will  strike  a  horizontal  floor 
simultaneously.1 

1PThis  interesting  experiment  was  suggested  by  Professor  F.  E.  Kester 
of  the  University  of  Kansas. 


CHAP.    XVII] 


FORCE  OF  MOTION 


289 


Problems 

1.  A  body  is  projected  horizontally  with  a  speed  of  40  feet  per  second. 
Where  will  it  be  at  the  end  of  one  second,  and  what  will  be  the  direction  and 
magnitude  of  its  resultant  velocity? 

Ans.     x  =  40  ft.;  y  =  -16.09  ft.;  v  =  51.4  ft.  per  sec.  at  an  angle  of  38° 
49'  with  the  horizontal. 

2.  A  train  is  running  over  a  trestle  with  a  velocity  of  40  feet  per  second. 
A  lump  of  coal  falls  from  the  tender  and  strikes  the  ground  30  feet  below. 
Find  its  resultant  velocity  by  means  of  the  energy  equations. 

Ans.     v  =  59.4  ft.   per  sec. 


b 


c 


\ 


\ 


\ 


FIG.  246. 

3.  A  train  is  running  over  a  trestle  with  a  velocity  of  40  feet  per  second. 
A  lump  of  coal  is  thrown  horizontally  at  right  angles  to  the  direction  of  the 
track  with  a  velocity  of  30  feet  per  second.     Find  the  direction  and  magni- 
tude of  the  velocity  after  2  seconds. 

Ans.     v  =  81.5  ft.  per  sec.  at  52°  09'  with  the  horizontal. 

4.  A  body  is  thrown  upward  at  an  angle  of  30  degrees  with  the  horizontal 
with  a  velocity  of  140  feet  per  second.     Find  the  horizontal  component  and 
the  vertical  component  of  its  velocity  after  2  seconds. 

Ans.    t'z  =  121;  24  ft.  per  sec.  vv  =  5.65  ft.  per  sec.  upward. 
19 


290  MECHANICS  [ART.  173 

5.  In  Problem  4,  how  long  will  the  body  continue  to  rise?     How  high  will 
it  go?     Solve  for  the  height  by  means  of  the  average  velocity. 

Ans.     2.176  sec.;   76.16  ft. 

6.  In  what  time  will  the  body  of  Problem  4  return  to  the  ground?     How 
far  from  the  starting  point  will  it  strike?  Ans.     x  =  527.6. 

If  a  projectile  is  thrown  upward  at  an  angle  a  with  the  hori- 
zontal with  a  velocity  of  v  feet  per  second,  the  components  of 
its  velocity  after  t  seconds  are 

Vx    =   VQ  COS  a,  (1) 

vy  =  VQ  sin  a  —  gt.  (2) 

The  time  required  to  reach  the  highest  point  in  the  path  is  found 
by  equating  the  vertical  component  to  zero.     This  method  is 
equivalent  to  treating  the  vertical  component  as  the  velocity  of  a 
body  moving  vertically  upward. 
To  find  the  position  at  any  time  t, 

x  =  (v0  cos  or)  t,  (3) 

at2 

y  =  (VQ  sin  a)t  -  ~  (4) 

The  first  term  of  Equation  (4)  gives  the  distance  the  body  would 
rise  with  a  constant  velocity  VQ  sin  a.  The  last  term  gives  the 
distance  it  would  fall.  The  difference  of  the  terms  gives  the  dis- 
tance above  the  horizontal  line  through  the  starting  point. 

Problems 

7.  Eliminate  t  from  Equations  (3)  and  (4)  and  derive  the  equation  of  the 
path  of  the  projectile  in  terms  of  x  and  y. 

8.  Derive   the  expression  for  the   horizontal  range  of  the   projectile. 

v  o  sin  2a 

Ans.     x  =  — • 

9 

9.  What  is  the  range  of  a  projectile  if  the  initial  velocity  is  160  feet  per 
second  at  an  angle  of  35  degrees  with  the  horizontal? 

Ans.     x  =  747.7.  ft. 

10.  Show  that  the  range  is  a  maximum  if  the  initial  velocity  makes  an 
angle  of  45  degrees  with  the  horizontal. 

11.  If  the  resistance  of  the  air  may  be  neglected,  what  is  the  maximum 
distance  which  a  basball  may  be  thrown  with  an  initial  velocity  of  100  feet 
per   second?  Ans.     310.8   ft, 

12.  A  16-pound  shot  leaves  an  athlete's  hand  at  a  distance  of  7  feet 
above  the  ground  with  a  velocity  of  36  feet  per  second.     How  far  will  it 
go,  if  thrown  at  45  degrees  with  the  horizontal?  Ans.  46.36  ft. 

13.  The  shot  of  Problem  12  moves  a  distance  of  7  feet  before  leaving  the 
athlete's  hand.     What  is  the  average  resultant  force.     Solve  by  work  and 
energy,  Ans.     57.3  Ib. 


CHAP.  XVII]  FORCE  OF  MOTION  291 

174.  Summary. — Newton's  Laws  of  Motion  are: 

1.  Every  body  continues  in  a  state  of  rest  or  of  uniform  motion 
in  a  straight  line,  unless  impelled  by  some  external  force  to 
change  that  condition. 

2.  Change  of  momentum  [Momentum  is  the  product  of  the 
mass  and  velocity]  is  proportional  to  the  applied  force  and  takes 
place  in  the  direction  in  which  the  force  acts. 

3.  For  every  action  there  is  an  equal  and  opposite  reaction. 
A  force  which  changes  the  motion  of  a  body  must  be  exerted 

by  some  outside  body.  The  attraction  of  the  Earth  for  a  projec- 
tile changes  the  motion  of  the  projectile.  The  attraction  of  the 
projectile  for  the  Earth  changes  the  motion  of  the  Earth.  Re- 
garded as  a  single  system  the  motion  of  the  center  of  mass  of 
the  projectile  and  the  Earth  is  not  changed. 

The  term  " applied  force"  means  the  resultant,  unbalanced, 
effective  force. 

Newton's  Second  Law  is  expressed  algebraically  by  the  formula, 

P  .=  — ,  Formula  XVII 

g 

in  which  the  force  P  is  expressed  in  terms  of  the  weight  of  unit 
mass.  When  the  time  is  expressed  in  seconds  and  the  distance 
in  feet  so  that  the  acceleration  is  in  feet  per  second  per  second, 
then  g  =  32.174.  If  the  distance  is  given  in  centimeters,  the 
value  of  g  is  981. 

Acceleration  is  the  rate  of  change  of  velocity. 


t  t  fit  /7/2 

&2  &  1  ^vv  \MV 

X2  —  x\  _  dx 

Vx    —    ~~, j ==   ~T7* 

c2  —  ti        dt 

The  displacement  is  the  product  of  the  average  velocity  multi- 
plied by  the  time.  When  the  acceleration  is  constant,  the  dis- 
placement may  be  represented  by  the  area  of  a  trapezoid  whose 
base  is  the  time  and  whose  terminal  ordinates  are  the  initial  and 
final  velocities.  The  average  velocity,  when  the  acceleration  is 
constant,  is  the  velocity  at  the  middle  of  the  time  interval. 

Kinetic  energy  is  the  energy  of  motion.     Its  equation  is 

U  =  —•  Formula  XXIV 

20 


292  MECHANICS  [ART.   175 

Potential  energy  is  the  energy  of  position.  The  sum  of  the 
potential  and  kinetic  energies  of  a  system  remains  constant  unless 
work  is  done  by  outside  bodies. 

Displacement,  velocity,  and  acceleration  are  vectors.  Work 
and  energy  are  not  vectors. 

When  a  body  is  moving  in  a  circle  with  uniform  speed  v,  the 
acceleration  toward  the  center  due  to  the  change  in  direction  is 

v2 

a  —  — 
r 

If  a  projectile  has  a  low  velocity  and  considerable  density,  so 
that  the  influence  of  the  air  resistance  is  negligible,  the  horizontal 
component  of  the  velocity  will  remain  unchanged  throughout 
the  flight  and  the  vertical  component  will  be  the  initial  vertical 
component  minus  the  change  due  to  gravity. 

Formula  XXII  applies  to  an  entire  connected  system  or  to 
any  part  of  such  system. 

175.  Miscellaneous  Problems 

1.  A  mass  of  20  pounds  slides  down  a  plane,  which  is  60  feet  long  and 
makes  an  angle  of  20  degrees  with  the  horizontal.     After  leaving  the 
inclined  plane,  it  slides  on  a  horizontal  plane.     The  coefficient  of  friction  of 
the  inclined  plane  is  0.2  and  the  coefficient  of  friction  of  the  horizontal 
plane  is  0.25.     How  far  will  the  body  go  on  the  horizontal  plane?  Solve  by 
work  and  energy.     What  is  the  acceleration  on  each  plane? 

Ans.     36.97  ft.;  4.96  ft./sec2;  -8.04  ft./sec.2 

2.  A  man  standing  on  the  ground  throws  a  ball  weighing  1  pound  with 
a  horizontal  velocity  of  60  feet  per  second.     What  energy  does  he  give  to  the 

ball? 


Ans.     V-  =  55.95  ft,lb. 

g 

3.  The  man  of  Problem  2  stands  on  a  car  which  is  moving  at  the  rate  of 
40  feet  per  second  and  throws  the  1-pound  ball  with  a  velocity  of  60  feet 
per  second  relative  to  the  car  in  the  direction  the  car  is  moving.     How  much 
energy  does  he  give  to  the  ball?     He  exerts  the  same  force  as  when  he  was 
standing  on  the  ground.     Explain  the  difference. 

Ans.     Energy  change  =  130.53  ft.-lb. 

4.  The  man  of  Problem  2  stands  on  a  car  which  is  moving  with  a  speed 
of  40  feet  per  second  and  throws  a  ball  weighing  1  pound  with  a  velocity  of 
60  feet  per  second  in  a  horizontal  direction  at  right  angles  to  the  direction 
of  the  car's  motion.     How  much  energy  does  he  impart  to  the  ball?     What 
is  the  total  energy  after  the  ball  has  fallen  20  feet?     What  is  the  direction 
of  the  motion  after  the  ball  has  fallen  20  feet? 

Ans.     55.95  ft.-lb.;  100.76  ft.-lb.  26°  27'  with  the  horizontal  in  a  vertical 
plane  which  makes  an  angle  56°  19'  with  the  direction  of  motion  of  the  car. 

5.  A  rope  is  thrown  over  a  cylinder  whose  axis  is  horizontal.     A  4-pound 
mass  is  hung  on  one  end  of  the  rope  and  a  1-pound  mass  is  hung  on  the  other 


CHAP.  XVII]  FORCE  OF  MOTION  293 

end.     The  coefficient  of  friction  is  0.2.     Find  the  tension  on  each  end  of 
the  rope  and  the  acceleration. 

6.  A  40-pound  mass  has  its  velocity  changed  during  4  seconds  from  120 
feet  per  second  east  to  40  feet  per  second  east.     Find  the  force  required. 

7.  A  40-pound  mass  has  its  velocity  changed  from  120  feet  per  second 
east  to  40  feet  per  second  east  while  it  goes  240  feet.     Find  the  acceleration 
by  means  of  the  average  velocity  and  find  the  force  required  to  change  the 
motion.     Solve  also  for  the  force  by  means  of  work  and  energy. 

8.  A  40-pound  mass  has  its  velocity  changed  from  120  feet  per  second 
east  to  40  feet  per  second  west  during  an  interval  of  5  seconds.     How  far 
does  it  travel  during  the  5  seconds  and  how  far  is  the  final  position  from  the 
initial  position?     Draw  the  diagram  of  velocity  and  time,  and  check  the 
distances  by  means  of  the  areas. 

9.  A  mass  of  60  pounds  has  its  velocity  changed  from  80  feet  per  second 
east  to  60  feet  per  second  north  in  an  interval  of  3  seconds.     Find  the 
direction  and  magnitude  of  the  resultant  force. 

10.  A  body  slides  down  an  inclined  plane  and  out  upon  a  horizontal 
plane  as  in  Problem  1.     If  the  coefficient  of  friction  is  the  same  for  both 
planes,  and  if  no  additional  energy  is  lost  when  the  direction  of  the  motion 
is  changed,  show  that  the  final  position  will  be  the  same  no  matter  what  is 
the  angle  of  the  inclined  plane.     Show  also  that  the  horizontal  distance  of 
the  final  position  from  the  initial  position  is  equal  to  the  initial  height 
multiplied  by  the  cotangent  of  the  angle  of  friction. 


CHAPTER  XVIII 
SYSTEMS  OF  UNITS 

176.  Gravitational  System. — The  system  of  units  used  in  the 
preceding  chapter  may  be  called  the  gravitational  system.     The 
unit  force  in  this  system  is  the  weight  of  unit  mass.     With 
English  units  the  unit  of  mass  is  the  pound  and  the  unit  of  force 
is  the  weight  of  one  pound  mass, — the  pound  mass  is  the  unit  of 
mass  and  the  pound  weight  is  the  unit  of  force.     With  metric 
units  the  unit  of  mass  is  the  gram  or  the  kilogram  and  the  unit 
of  force  is  the  weight  of  a  gram  mass  or  of  a  kilogram  mass. 
These  are  the  ordinary  units  in  every  day  use.     They  have  the 
apparent   disadvantage   that   the   weight   of  unit   mass   varies 
slightly  with  latitude  and  altitude.     The  standard  unit  of  force 
is  the  weight  of  the  unit  of  mass  at  45  degrees  latitude  at  the  sea 
level. 

In  the  gravitational  system,  the  constant  in  the  equation  which 
expresses  the  relation  of  force,  mass  and  acceleration  is  g. 

P  =  — ;  Formula  XXII 

9 

TYIV^ 

kinetic  energy  =  - — .  Formula  XXIV 

^9 

177.  Absolute  Systems. — In  the  absolute  systems,  the  constant 
in  the  equation  of  force,  mass,  and  acceleration  is  unity.     The 
equation  reads 

P  =  ma  Formula  XXV 

If  m  =  1  and  a  =  1  in  Formula  XXV,  then  the  formula  defines 
P  as  the  force  which  gives  unit  acceleration  to  unit  mass.  The 
equation  gives  a  definition  of  the  unit  of  force  which  is  everywhere 
constant. 

Since  force  is  the  product  of  mass  multiplied  by  acceleration, 
and  since  the  dimensional  equation  of  acceleration  is  L77"2,  it 
follows  that  the  dimensional  equation  of  force  is  MLT~2. 

The  expression  for  kinetic  energy  in  an  absolute  system  is 

TT  -  H! 

T 

294 


CHAP.  -XVIII]  SYSTEMS  OF  UNITS  295 

The  dimensional  equation  of  kinetic  energy  is  ML2T~2.  The 
dimensional  equation  of  work  is  the  same  as  that  of  kinetic 
energy. 

178.  The  Centimeter-Gram-Second  System. — An  absolute 
system  in  which  the  unit  of  mass  is  the  gram  and  the  unit  of 
acceleration  is  the  centimeter  per  second  per  second  is  called  the 
centimeter-gram-second  system.  The  name  is  abbreviated  to 
the  C.G.S.  system. 

In  this  system,  the  unit  of  force  is  the  dyne.  The  dyne  is  that 
force  which  gives  to  a  gram  mass  an  acceleration  of  1  centimeter 
per  second  per  second.  The  weight  of  a  gram  mass  gives  to  a  gram 
an  acceleration  of  g  centimeters  per  second  per  second.  The 
value  of  g  varies  slightly.  For  most  places  it  is  between  980  and 
981.  The  weight  of  a  gram  is  about  981  dynes.  In  the  C.G.S. 
system,  weight  =  mg;  in  the  gravitational  system,  weight  =  m. 

The  unit  of  energy  in  the  C.G.S  system  is  called  the  erg.  An 
erg  is  the  work  done  by  a  force  of  one  dyne  when  the  displace- 
ment is  1  centimeter.  Ergs  are  reduced  to  gram-centimeters  of 
work  by  dividing  by  981. 

The  C.G.S  system  is  used  by  physicists.  It  has  the  advantage 
of  apparent  simplicity  and  it  gives  a  unit  of  force  which  does  not 
change  with  locality.  The  commercial  electrical  units  were 
originally  based  on  the  C.G.S  system.  The  official  definitions 
now  used  do  not,  however,  depend  directly  upon  this  system. 


Problems 

1.  A  mass  of  20  grams  has  its  velocity  changed  from  60  centimeters 
per  second  to  20  centimeters  per  second  in  4  seconds.     Find  the  effective 
force  in  dynes  and  the  change  in  kinetic  energy. 

Ans.     P  =  200  dynes;  change  in  energy  =  32,000  ergs. 

2.  Calculate  the  displacement  in  Problem  1  by  means  of  the  average 
velocity  and  compute  the  work  done  by  the  force. 

3.  A  mass  of  50  grams  is  placed  on  one  side  of  an  Atwood  machine  and  a 
mass  of  30  grams  on  the  other      Find  the  acceleration  and  the  tension  in 
the  cord,  if  g  is  980.6  at  that  locality. 

4.  What  is  the  kinetic  energy  in  ergs  of  a  mass  of  2  kilograms  which  is 
moving  with  a  velocity  of  4  meters  per  second.     .4ns.     U  =  16  X  107  ergs. 

5.  A  mass  of  20  kilograms  is  moving  on  a  horizontal  plane  with  a  velocity 
of  5  meters  per  second.     It  is  brought  to  rest  in  10  meters  by  the  friction 
of    the    plane.     Find    the    coefficient    of    friction.         Ans.    f  =  0.127. 

6.  How  many  dynes  are  equal  to  a  standard  pound  force? 

Ans.     444,820    dynes  =  1    standard    pound. 


296  MECHANICS  [Awr.   179 

7.  Reduce  a  standard  foot-pound  to  ergs. 

Ans.     13,558  X  103    ergs  =  1    standard    foot-pound. 

8.  How  many  dynes  are  equal  to  the  weight  of  one  gram  at  45  degrees 
latitude  at  the  sea  level?  Ans.     980.66  dynes. 

179.  The  Foot-Pound-Second  System.— The  Foot-Pound- 
Second  System  is  an  absolute  system  in  which  the  pound  is  the 
unit  of  mass  and  the  foot  per  second  per  second  is  the  unit  of 
acceleration.  The  unit  of  force  is  called  the  poundal.  A  poundal 
is  that  force  which  gives  to  a  pound  mass  an  acceleration  of  1  foot 
per  second  per  second.  Since  the  weight  of  a  pound  mass  gives 
to  a  pound  mass  an  acceleration  of  g  feet  per  second  per  second, 
it  is  evident  that  a  pound  force  is  equal  to  g  poundals  force. 
A  standard  pound  force  is  32.174  poundals.  A  poundal  is  a 
little  less  than  the  weight  of  one-half  ounce.  The  unit  of  work 
and  energy  in  this  system  is  the  foot-poundal. 

Problems 

1.  A  mass  of  60  pounds  is  moving  with  a  velocity  of  20  feet  per  second. 
Find  the  force  in  poundals  which  will  stop  it  in  a  space  of  20  feet. 

Ans.     P  =  600  poundals. 

2.  A  car  weighing  2400  pounds  is  moving  with  a  velocity  of  30  feet  per 
second.     Find  its  kinetic  energy  in  foot-poundals. 

Ans.     108  X  104  foot-poundals. 

3.  A  40-pound  mass  on  a  20-degree  inclined  plane  is  attached  to  a  rope 
which  runs  up  the  plane,  passes  over  a  smooth  pulley,  and  supports  a  mass 
of  24  pounds.     Find  the  acceleration  and  the  tension  on  the  rope  in  poundals. 

In  problem  3,  and  in  all  problems  in  which  a  force  depends  upon 
the  weight  of  some  body,  it  is  necessary  to  know  the  local  value 
of  g  in  order  to  get  the  true  results  in  poundals.  If  an  assumed 
value  of  g  is  used,  the  results  will  be  in  error  in  the  ratio  of  the 
assumed  value  of  g  to  the  true  local  value.  When  the  Formulas 
of  Art  165  are  employed  with  the  standard  value  of  g,  the  accel- 
eration of  Problem  3  will  be  in  error  in  the  ratio  of  the  standard 
g  to  the  local  g.  The  tension,  on  the  other  hand,  will  be  correct 
in  local  pounds.  As  far  as  the  absolute  accuracy  of  the  results 
are  concerned,  one  system  has  no  advantage  over  the  other. 

Forces  are  usually  measured  by  means  of  the  local  weight  of  a 
copy  of  a  standard  mass.  When  it  is  desired  to  calibrate  a  spring 
balance,  standard  " weights"  are  hung  on  it.  (Methods  for 
calibrating  a  spring  directly  in  absolute  units  are  given  in  Chapter 


CHAP.  XVIII]  SYSTEMS  OF  UNITS  297 

XIX.)  Spring  balances  are  generally  graduated  to  read  in  pounds 
and  not  in  poundals. 

In  poundals,  P  =  ma;  Formula  XXV 

,     „       poundals       ma  ^          ,    VVTT 

in  pounds,  P  =  -     = Formula  XXII 

Q  Q 

For  the  student  who  has  begun  with  the  absolute  systems, 
Formula  XXII  may  be  regarded  as  obtained  from  Formula  XXV 
by  division  by  g. 

180.  The  Engineer's  Unit  Mass. — A  force  of  one  pound  gives 
to  a  mass  of  one  pound  an  acceleration  of  32.17  feet  per  second, 
per  second.  A  force  of  one  pound  gives  to  a  mass  of  32.17  pounds 
an  acceleration  of  1  foot  per  second  per  second.  A  mass  of  32.17 
pounds  is  called  the  engineer's  unit  of  mass.  It  seems  to  be  as- 
sumed that  it  is  desirable  to  express  Newton's  Second  Law  in 
the  form  of 

Force  =  mass  X  acceleration  (1) 

When  the  force  in  Equation  (1)  is  expressed  in  pounds,  it  is 
necessary  to  have  this  large  unit  of  mass.  This  unit  of  mass  until 
recently  had  no  name  except  " engineer's  unit,"  or  simply  the 
''unit  of  mass."  Professor  E.  R.  Maurer  has  suggested  the  name 
of  gee-pound,  since  the  unit  is  g  times  as  large  as  the  pound  unit. 
(Some  British  writers  call  it  the  slugg.)  To  find  the  number 
of  gee-pounds  in  a  given  mass,  divide  the  mass  in  pounds  by  g. 
Some  engineering  writers  express  the  number  of  pounds  mass  by 

W 

W.     The  number  of  gee-pounds  is  — .     If  M  is  the  mass  in 

y 

W 

gee-pounds,  M  =  — .     Newton's  Second  Law  is  written, 
y 

W 

Force  in  pounds  =  —a.  (2) 

ij 

Since  W  is  really  the  mass  in  pounds  (though  it  is  often  called 
the  weight) 

Wa       ma 


and  Equation  (2)  is  equivalent  to  Formula  XXII.     The  dif- 
ference  between   Equation    (2)    as   used   by   some  writers  and 

W 
Formula  XXII  lies  in  the  fact  that  —  is  defined  as  being  the  mass 


298  MECHANICS  [ART.  181 

of  the  body  in  a  new  unit,  while  in  Formula  XXII,  W  or  m  repre- 

sents the  mass  in  pounds,  and  -  is  the  ratio  of  the  acceleration 

y 

of  the  body  to  the  acceleration  of  gravity,  or  g  is  regarded  as  a 
constant  occurring  in  the  formula. 

It  is  sometimes  stated  that  W  in  this  equation  is  the  weight 
of  the  mass  on  a  standard  spring  balance.  A  body  may  weigh 
332  pounds  at  one  locality  and  321.5  pounds  at  another.  If 
g  at  the  first  locality  is  32.2  and  at  the  second  locality  is  32.15, 
the  mass  is  found  to  be  10  units  in  each  case.  The  formula 
P  =  Ma  gives  the  correct  accelerating  force  in  standard  pounds, 

Mv2 
and  the  formula  U  =  —  ~—  gives  the  correct  energy  in  standard 

foot-pounds.  Unfortunately,  the  use  of  such  a  standard  spring 
balance  assumes  that  the  true  value  of  g  is  known.  The  same 
errors,  therefore,  come  in  as  with  the  use  of  other  methods. 
Moreover,  such  standard  spring  balances  do  not  exist,  and  would 
not  be  sufficiently  accurate  for  refined  experiments  if  they  were 
made.  When  it  is  necessary  to  measure  forces  with  an  accuracy 
greater  than  the  variation  of  gravity,  beam  balances  are  employed 
and  the  value  of  g  is  determined  by  vibration  experiments. 

Wa 
The  formula  P  =  -  -  may  be  written, 


In  this  equation  W  is  regarded  as  a  force.  The  effective  force  is  to 
the  weight  of  the  body  as  the  acceleration  produced  by  the  force 
is  to  the  acceleration  produced  by  the  weight  of  the  body.  Equa- 
tion (3)  may  be  regarded  as  the  simplest  form  of  the  equation  of 
acceleration.  It  does  not  involve  any  unusual  unit,  such  as  the 
poundal  force  or  the  gee-pound  mass.  If  only  linear  accelera- 
tion were  to  be  considered,  Equation  (3)  would  be  the  best 
form  to  use.  On  account  of  the  necessity  of  applying  the 
formulas  to  angular  acceleration  and  to  problems  of  kinetic 
energy,  it  is  advisable  to  emphasize  the  idea  of  the  mass. 
Formula  XXII  is,  therefore,  best  adapted  for  general  use. 

181.  Summary.  —  Systems  of  units  depend  upon  the  value  of 

the  constant  k  in  the  equation  P  =  -  r~.     The  constant  is  unity 


CHAP.  XVIII]  SYSTEMS  OF  UNITS  299 

in  the  absolute  systems  and  the  equations  for  acceleration  and 
kinetic  energy  are  written 

P  =  ma;  Formula  XXV 


In  these  systems,  the  units  of  mass,  length,  and  time  are  arbi- 
trarily defined  and  the  unit  of  force  is  derived  from  Formula  XXV. 
The  results  are  changed  into  gravitational  units  by  dividing  by  g. 
In  the  C.G.S.  system,  the  dyne  is  the  unit  of  force.  A  dyne  is 
the  force  which  gives  to  a  gram  mass  an  acceleration  of  1  centi- 
meter per  second  per  second.  In  the  foot-pound-second  system, 
the  poundal  is  the  unit  of  force.  A  poundal  is  the  force  which 
gives  to  a  pound  mass  an  acceleration  of  1  foot  per  second.  One 
pound  force  is  equal  to  32.174  poundals.  One  gram  force  is 
equal  to  981  dynes. 

In  the  so-called  Engineer's  System,  the  units  of  time  and 
length  are  the  same  as  in  the  other  systems.  The  unit  offeree  is 
taken  arbitrarily  and  the  unit  of  mass  is  defined  by  Formula  XXV. 
Standard  units  of  force  and  mass,  as  required  by  this  system, 
do  not  exist.  Physicists  use  the  pound  as  the  unit  of  mass  and 
define  the  unit  of  force  by  Formula  XXV.  Engineers,  also, 
use  the  pound  as  the  unit  of  mass  and  define  the  unit  of  force  as 
the  weight  of  a  pound  mass  at  the  standard  position.  Since 
nothing  is  gained  by  the  use  of  the  so-called  Engineer's  Unit 
of  mass,  it  might  well  be  dropped  from  technical  literature. 

Many  books  have  the  formulas, 

Wa 
Force  in  pounds  =  -  ; 

Wv2 
Kinetic  energy  in  foot-pounds  =  —=  — 

These  are  identical  with  the  formulas  of  the  preceding  chapter 
except  that  W  is  used  instead  of  m  for  the  mass  in  pounds.  The 
student  is  advised  to  think  of  g  as  a  mere  coefficient  occurring 
in  the  formula  on  account  of  the  units  employed,  and  not  to 

W 

consider  —  as  representing  the  mass. 

\y 

The  absolute  systems  have  their  uses  in  some  scientific  work. 
In  problems  where  forces  depend  upon  the  weight  of  some  bodies, 
it  is  necessary  to  know  the  local  value  of  g.  In  these  systems, 


300 


MECHANICS 


[ART.    181 


weight  =  mg.  While  these  systems  eliminate  g  from  the  formula 
for  acceleration,  they  insert  g  into  many  problems  of  statics. 

Engineers  are  advised  to  learn  Formulas  XXII  and  XXIV  as 
given  in  the  preceding  chapter.  These  formulas  are  apparently 
more  complex  than  those  of  the  other  systems.  It  is  easier, 
however,  for  the  engineer  who  is  not  working  daily  with  such  ideas 
to  remember  (or  look  up)  an  equation  with  an  additional  letter 
in  it  than  to  remember  to  divide  by  g  to  transform  to  practical 
units. 

The  systems  of  units  are  given  in  Table   1  below. 

TABLE  I. — SYSTEMS  OF  UNITS 


System 

Force 

Mass 

Length 

Weight 

Formulas 

Gravitational  . 

Pound 

Pound 

Foot 

m 

ma              mv2 
"    9    '        =  2g 

C.G.S.... 

Dyne 

Gram 

Cm. 

mg 

P  —  ma'  U  — 

2 

Ft.  Lb.;  8.... 

Poundal 

Pound 

Ft. 

mg 

mv2 
P  =  ma',  U  =  -£- 

Engineer's  

Pound 

Gee-pound 

Ft. 

mg 

Tr       mv2 
P  =  ma',  U  =  ~2~ 

CHAPTER  XIX 


FORCE  WHICH  VARIES  AS  THE  DISPLACEMENT 

182.  The  Force  of  a  Spring. — The  deformation  of  a  spring  or 
other  elastic  body  is  proportional  to  the  force.  Figure  247,  I, 
shows  a  helical  spring  which  is  not  loaded.  A  load  of  1  pound 
stretches  the  spring  a  definite  amount  d  as  shown  in  Fig.  247,  II. 
A  load  of  2  pounds  would  produce  an  elongation  of  2d.  The 
forces  on  the  spring  are  downward  and  the  elongation  is  down- 
ward. The  force  with  which  the  spring  acts  on  the  loads  is 
upward  opposite  to  the  deformation. 


O 


FIG.  247. 


When  a  body  is  attached  to  a  spring,  there  is  some  position 
at  which  it  remains  at  rest.  In  this  position,  either  the  force  of 
the  spring  is  zero,  or  the  resultant  of  the  force  of  the  spring  and 
the  other  forces  is  zero.  Figure  248  represents  a  body  on  a  flat 
strap  spring.  The  plane  of  the  paper  is  supposed  to  be  horizontal 
in  Fig.  248,  I.  The  weight  of  the  body  tends  to  bend  the  spring 
in  a  vertical  direction  but  not  in  a  horizontal  direction.  Equi- 
librium exists,  therefore,  when  the  spring  viewed  from  above, 
appears  straight.  If  the  body  is  displaced  toward  the  right,  as 

301 


302 


MECHANICS 


[ART.    182 


shown  in  Fig.  248,  III,  the  spring  tends  to  move  it  backward 
toward  the  left.  If  it  is  displaced  toward  the  left  from  the 
position  of  equilibrium,  the  spring  tends  to  move  it  toward  the 
right. 

If  the  spring  in  Fig.  248  exerts  a  force  of  K  pounds  when  it  is 
deformed  a  distance  of  1  foot,  it  will  exert  a  force  of  2K  pounds 

when  it  is  deformed  a  distance  of  2 
feet  and  a  force  of  Kx  pounds  when 
it  is  deformed  a  distance  of  x  feet. 
The  force  acts  in  a  dipeclion  opposite 
to  the  displacement.  The  direction 
and  magnitude  of  the  force  exerted  by 
the  spring  on  the  body  is  given  mathe- 
matically by  the  equation. 

P  =  -Kx.  (1) 

In  Equation  (1)  K  is  the  force  of  the 
spring  when  the  displacement  is 
unity,  x  is  the  displacement  from  the 
position  of  equilibrium,  and  P  is  the 
force  of  the  spring  when  the  displace- 
ment is  x.  The  negative  sign  indi- 
cates that  the  direction  of  the  force 
which  the  spring  exerts  on  the 
body  is  opposite  the  direction  of  the 
displacement. 

Equation  (1)  applies  also  when  a 
body  is  subjected  to  the  force  of  a 
spring  and  a  constant  force.  The  dis- 
placement is  measured  from  the  posi- 
tion of  equilibrium  at  which  the  force 
of  the  spring  is  equal  and  opposite  to  the  constant  force.  Figure 
249  shows  a  vertical  helical  spring.  It  is  assumed  that  a  load  of  24 
pounds  stretches  this  spring  a  distance  of  3  feet,  as  shown  in  Fig. 
249,  II.  The  constant  K  is  8  pounds.  In  Fig.  249,  II,  the  mass 
of  24  pounds  is  subjected  to  a  downward  pull  of  24  pounds  which 
is  due  to  gravity,  and  to  an  upward  pull  of  24  pounds  which  is  due 
to  the  reaction  of  the  spring.  If  the  24-pound  mass  is  now 
displaced  an  additional  foot  downward,  the  elongation  of  the 
spring  becomes  4  feet  and  the  force  in  it  becomes  32  pounds. 
The  force  of  gravity  remains  24  pounds.  The  resultant  force 


in 


FIG.  249. 


CHAP.  XIX]     FORCE  VARIES  AS  DISPLACEMENT          303 

on  the  24-pound  mass  is  an  upward  force  of  8  pounds.  If  the 
24-pound  mass  is  displaced  a  distance  y  from  the  position  of 
equilibrium,  the  total  elongation  of  the  spring  is  3  +  y  feet,  and 
the  upward  pull  of  the  spring  becomes  24  +  Sy  pounds.  The 
downward  pull  of  gravity  is  still  24  pounds..  The  resultant 
force  on  the  body,  when  the  displacement  is  —  y  feet,  is  equal  to 
Sy  pounds.  If  the  24-pound  mass  is  displaced  a  distance  y 
upward  from  the  position  of  equilibrium,  the  total  elongation  of 
the  spring  is  3  —  y  feet  and  the  pull  of  the  spring  is  24  —  Sy, 
provided  y  is  not  more  than  3  feet.  The  downward  pull  of  the 
earth  is  still  24  pounds  so  that  the  resultant  force  on  the  body  is 
—  Sy  pounds  when  the  displacement  is  -\-y  feet  from  the  position 
of  equilibrium. 

If  the  origin  of  coordinates  is  taken  at  the  position  of  equi- 
librium under  the  action  of  the  constant  force  and  the  reaction 
of  the  spring,  the  expression  for  the  resultant  force  on  the  body  is 
P  =  -KxorP  =  -Ky. 

If  the  displacement  in  Fig.  249  is  greater  than  3  feet,  the  spring 
will  come  into  compression.  If  the  force  required  to  compress 
the  spring  unit  distance  is  the  same  as  the  force  required  to 
stretch  it  unit  distance,  Equation  (1)  will  remain  valid.  A  long 
helical  spring  such  as  is  shown  in  Fig.  249  will  buckle  in  compres- 
sion so  that  the  force  required  to  compress  it  unit  distance  is 
not  the  same  as  the  force  which  stretches  it  unit  distance.  With 
such  a  spring  Equation  (1)  will  not  hold  when  it  is  shortened 
below  its  natural  length. 

Problems 

1.  A  spring  board  is  deflected  5  inches  downward  by  a  load  of  100  pounds 
on  the  free  end.     What  is  the  value  of  K?  Ans.     K  =  240  Ib. 

2.  What  is  the  deflection  of  the  spring  board  of  Problem  1  when  a  boy 
weighing  120  pounds  stands  on  the  end?  Ans.     0.5  ft. 

3.  In  Problem  2,  what  is  the  resultant  force  on  the  boy  when  the  spring 
board  is  deflected  8  inches  below  the  position  at  which  it  stands  when  not 
loaded.  Ans.     40  Ib.  upward. 

4.  In  Problem  2,  what  is  the  resultant  force  on  the  boy  when  the  spring 
is  deflected  3  inches  downward  from  the  position  of  equilibrium  under  no 
load?  Ans.     60  Ib.  downward. 

5.  Solve  Problem  4  if  the  deflection  is  1  inch  upward  from  the  position 
of  equilibrium  under  no  load  and  the  boy  is  not  fastened  to  the  spring 
board.  Ans.     P  =  - 120  Ib. 

6.  A  body  is  placed  on  a  horizontal  plane  and  attached  to  one  end  of  a 
spring.     The  natural  length  of  the  spring  is  3  feet  and  a  pull  of  12  pounds 


304 


MECHANICS 


[ART.    183 


stretches  it  6  inches.  A  cord  attached  to  the  body  runs  horizontally  over 
a  smooth  pulley  and  supports  a  mass  of  60  pounds,  as  shown  in  Fig.  250,  II. 
What  is  the  zero  position  of  the  right  end  of  the  spring  when  the  system  is 
in  equilibrium?  When  the  spring  is  stretched  to  a  length  of  7  feet,  what  is 
the  resultant  force  on  the  body? 


FIG.  250. 

7.  Figure  251  shows  two  springs  attached  to  fixed  points  A  and  C,  which 
are  at  the  same  level  and  are  10  feet  apart.  The  natural  length  of  each 
spring  is  3  feet  and  a  force  of  20  pounds  stretches  each  spring  1  foot.  A 
body  B  is  fastened  to  the  free  ends  of  these  springs.  What  is  the  resultant 
force  on  the  body  when  it  is  displaced  1  foot  from  the  position  of  equilibrium, 
and  when  it  is  displaced  x  feet  from  that  position?  Ans.  K  =  — 40x. 


8.  The  two  springs  of  Problem  7  are  replaced  by  a  single  spring  of  natural 
length  of  6  feet.  It  takes  a  force  of  10  pounds  to  stretch  this  spring  1  foot. 
A  body  is  attached  4  feet  from  the  left  end  of  the  stretched  spring.  What 
force  will  displace  this  body  1  foot  in  the  direction  of  the  length  of  the 
spring? 

183.  Potential  Energy  of  a  Spring. — The  work  done  by  a  force 
is  the  average  force  multiplied  by  the  displacement.  When 
the  force  varies  as  the  displacement,  the  average  force  is  the 
mean  of  the  initial  and  final  forces.  If  K  is  the  force  which 
produces  unit  deformation  in  a  spring  and  si  is  the  initial  defor- 
mation, the  initial  force  is  Ksi.  If  §2  is  the  final  deformation, 


CHAP.  XIX]     FORCE  VARIES  AS  DISPLACEMENT          305 
the  final  force  is  Ks2.     The  average  force  for  the  displacement 


s2  —  si  is 


K(sl  - 


Area  = 


Equation  (1)  gives  the  increase  of  potential  energy. 

Work  is  often  represented  by 
the  area  of  a  diagram  in  which 
force  is  the  ordinate,  and  defor- 
mation or  displacement  is  the 
abscissa.  Figure  252  is  the  work 
diagram  for  a  spring.  The  dia- 
gram is  a  trapezoid.  The  par- 
allel sides  are  Ksi  and  Ksz, 
and  the  distance  between  the  sides  is  s%  —  s\.  If  the  initial 
displacement  is  zero,  the  diagram  is  a  triangle  and  the  total 


h  —  5/  —  ^ 


.   . 
work  is 


Problems 


1.  In  Problem  1  of  Art.  182,  what  is  the  work  done  in  deflecting  the  board 
6  inches?     Solve  without  the  formula. 

2.  In  Problem  1  of  Art.  182,  what  is  the  work  done  in  deflecting  the 
board  1  foot?     What  is  the  work  of  the  last  6  inches  of  the  foot?     Con- 
struct the  work  diagram  to  the  scale  of  1  inch  =  50  pounds  and  1  inch  = 
0.25  ft. 


184.  Velocity  Produced  by  Elastic  Force.  —  When  the  force  is 
proportional  to  the  displacement,  as  in  the  case  of  a  spring,  the 
expression  for  the  force  is 


P  =  - 


(1) 


in  which  P  is  the  force  exerted  by  the  spring,  x  is  the  displace- 
ment, and  K  is  the  force  when  the  displacement  is  unity.  The 
sign  is  negative  for  a  spring  because  the  direction  of  the  force 
exerted  by  the  spring  is  opposite  the  direction  of  the  displacement. 
If  a  body  of  mass  m  is  attached  to  the  spring,  as  in  Fig.  253, 
the  acceleration  of  this  body  is  given  by  Formula  XXII. 


20 


-  Kx  =  — 
Q 


(2) 


306 


MECHANICS 


[ART.    184 


Since  acceleration  along  the  X  axis  is  ^p  Equation  (2)  may  be 
written 

/XM     /72<y« 

(3) 


m  cr: 

7^ 

d2z 


A  differential  equation  of  the  form  ^  =  a  function  of  x  is  solved 
by  first  multiplying  each  side  by  dx. 


m  dx  d2x 
-Jfe  &,-___=- 


m  dx  d2x 
~gdt~dt' 


dx          d2x 
Since   ^7  =  v,  -n-  =  dt;,  and  Equation  (2)  becomes 


dt 


m 


(4) 


(5) 


(6) 


In  Fig.  253,  I,  the  mass  m  is  at  its  position  of  equilibrium 
with  equal  tension  on  the  two  springs.     In  Fig.  253,  II,  the  body 


—  Kxdx  =  —  v  dv. 

g 

Integrating  Equation  (5), 


has  been  displaced  a  distance  r  toward  the  right  and  is  assumed 
to  be  stationary.  If  let  free  at  this  point,  the  body  will  move 
to  the  position  of  Fig.  253,  I,  and  will  continue  to  move  to  a 
distance  r  on  the  left  of  the  position  of  equilibrium.  When 
x  =  r,  v  =  0.  Substituting  in  Equation  (6), 

-  — 

1  ~          2 


A  substitution  for  Ci  in  Equation  (6)  gives 


mv* 


Kr*       Kx* 


Formula  XXVI 


At  the  position  of  maximum  deflection,  at  which  x  =  r  and 
the  velocity  is  zero,  the  body  is  at  positive  elongation.  The 
distance  r  is  called  the  amplitude. 


CHAP.  XIX]     FORCE  VARIES  AS  DISPLACEMENT          307 

The  first  member  of  Formula  XXVI  is  the  expression  for 
kinetic  energy,  consequently  the  second  member  must  also 

Kr2 
represent  energy.     The  term  —^-  gives  the  potential  energy 

Kx2 
when  the  displacement  is  r.     The  term  — ~-   gives  the  potential 

,                                                                         Kr2      Kx2 
energy  when  the  displacement  is  x.     The  expression  —= ^— 

Z  Zi 

represents  the  work  done  on  the  spring  when  the  displacements 
changed  from  x  to  r  or  the  work  done  by  the  spring  when  the 
displacement  is  changed  from  r  to  x. 

When  the  displacement  is  r  feet,  the  mass  m  is  stationary 
and  all  the  energy  is  in  the  form  of  potential  energy  of  the  spring. 
When  the  displacement  is  reduced  to  x  the  potential  energy  of 

Kx2  Kr2          Kx2 

the  spring  is  —FT~.     The  difference  between  —^r-  and  — ^—  repre- 
z  z  z 

sents  the  energy  which  is  given  up  by  the  spring  to  the  body 
and  is  transformed  into  kinetic  energy  of  the  mass.  When 
x  =  r,  all  the  energy  is  potential.  When  x  =  0,  all  the 
energy  is  kinetic.  When  x  =  —  r,  all  the  energy  is  again  poten- 
tial and  the  body  comes  to  rest  with  negative  elongation  equal 
to  the  positive  elongation.  This  is  the  well  known  case  of 
vibratory  motion.  The  total  motion  is  twice  the  amplitude. 
If  there  were  no  loss  of  energy,  the  vibration  would  continue 
indefinitely.  Since  there  is  always  some  loss  of  energy,  the 
amplitude  of  the  vibration  will  slowly  diminish  and  the  body  will 
come  to  rest  at  the  position  of  equilibrium  after  a  long  interval. 
Formula  XXVI  might  have  been  derived  directly  from  the 
relation  of  the  potential  to  the  kinetic  energy,  without  the  use 
of  integration.  The  integration  method  here  given  is  general, 
however,  and  may  be  applied  to  any  problem  in  which  the 
effective  force  is  a  function  of  the  displacement.  It  is  advisable, 
therefore,  to  use  the  method  in  this  case  where  it  may  be  checked 
in  order  that  it  may  be  employed  with  confidence  in  other  cases 
where  it  is  not  possible  to  verify  the  results. 

Problems 

1.  A  spring  suspended  from  a  fixed  support  is  elongated  3  feet  when  a 
24-pound  mass  is  hung  on  it.  The  spring  is  stretched  an  additional  2  feet 
downward  and  then  released.  What  will  be  the  velocity  when  the  spring 
is  stretched  1  foot  below  the  position  of  equilibrium?  What  will  be  the 
velocity  when  the  body  is  1  foot  above  the  position  of  equilibrium?  What 


308  MECHANICS  [ART.  185 

will  be  the  maximum  velocity  upward?     What  will  be  the  maximum  velocity 
downward?          Ans.    5.67  ft.  per  sec.;  5.67  ft.  per  sec.;  6.54  ft.  per  sec. 

2.  A  24-pound  mass  rests  on  a  horizontal  table  between  the  ends  of  two 
horizontal  springs  similar  to  those  of  Fig.  251.     It  requires  a  force  of  20 
pounds  to  move  the  body  1  foot  in  the  direction  of  the  length  of  the  springs. 
The  body  is  displaced  a  distance  of  3  feet  and  then  released.     If  there  is  no 
friction,  find  the  maximum  velocity  of  the  body,  the  velocity  when  the 
displacement  is  1  foot,  and  when  the  displacement  is  2  feet. 

Ans.     15.53  ft.  per  sec.;  14.65  ft.  per  sec.;  11.58  ft.  per  sec. 

3.  Solve  Problem  2  if  the  coefficient  of  friction  between  the  body  and  the 
table  is  0.1. 

When  the  displacement  is  2  feet,  the  energy  given  up  by  the  spring  is 
20(9  _  4\ 
—  *-=  -  '  =  50  ft.-lb.     The  work  done  in  moving  the  body  1  foot  against 

the  friction  is  2.4  ft.-lb.     The  kinetic  energy  is  50  —  2.4  =  47.6  ft.-lb. 
The  velocity  of  the  24-pound  mass  is  11.29  ft.  per  sec. 

The  maximum  velocity  is  at  the  point  at  which  the  force  of  the  spring 
is  equal  to  the  friction.     At  this  point,  the  resultant  force  is  zero  and  the 

2.4 
acceleration  is  reduced  to  zero.     This  position  is  sir  =  0-12  feet  from  the 

point  of  zero  tension  in  the  spring. 


=  ??  (9  -  0,0144)  -  2.4  X  2.88; 
max  v  =  14.91  ft.  per  sec. 

4.  In  Problem  3,  find  the  maximum  negative  elongation  from  the  position 
of  zero  force  in  the  spring.  Ans.  x  =  —2.76  ft. 

6.  A  mass  of  40  pounds  is  placed  on  a  small  platform  supported  by  a 
spring.  The  weight  of  the  40-pound  mass  elongates  the  spring  a  distance 
of  2  feet.  The  platform  and  mass  are  pulled  down  an  additional  3  feet  and 
then  released.  If  no  energy  is  lost  and  if  the  mass  is  not  fastened  to  the 
platform,  how  high  will  the  40  pounds  rise  above  the  position  of  equilibrium? 

Ans.  3.25  ft. 

185.  Vibration  from  Elastic  Force.—  From  Formula  XXVI, 
which  applies  to  the  force  of  a  spring,  or  to  any  force  which  is 
proportional  to  the  displacement  and  is  opposite  the  direction 
of  the  displacement,  the  velocity  is 

v*  =  ^  (r*  -  z2).  (1) 

In  this  equation,  v  is  the  velocity  when  the  displacement  is  x,  r 
is  the  amplitude  or  maximum  displacement,  and  K  is  the  force 
in  pounds  exerted  on  the  mass  m  when  the  displacement  is  1  foot. 


CHAP.  XIX]     FORCE  VARIES  AS  DISPLACEMENT 


309 


If  the  time  of  positive  elongation  at  which  x  =  r  is  taken  as  the 
zero  time, 

7T 


C2  =  sin"1  1;  sin  C2  =  1;  C2  =  «J 

\Kg  .      .  .     "x       TT       •          xt 

—  t  =  sin-1 =  cos-1  -, 

m  r       2  r 


cos 


x  =  r  cos 


^A 

m    I 


(4) 
(5) 
(6) 
(V) 


Equation  (6)  gives  the  displacement  of  the  mass  in  terms  of  the 
interval  which  has  elapsed  since  it  was  at  positive  elongation. 


When  x  =  r,   t  =  0.     When   x  =  0,  cos 


and 


' —  t  =  ~.     The  time  from  positive  elongation  to  zero  elongation 

tn          - 

is  one-half  the  time  of  single  vibration  and  one-fourth  the  time  of  a 
complete  period.  From  the  expression  above,  the  time  from 
positive  elongation  to  zero  elongation  is 


i-i 


m 

Kg 


(8) 


The  time  of  a  single  vibration  from  positive  elongation  to  negative 
elongation,  or  from   zero  elonga- 
tion   to    positive   elongation   and 
back  to  zero  elongation,  is  given 

by  the  equation 

• — 
m 

Kg 


£:-  (9) 


The  time  of  a  complete  period 
from  positive  elongation  to  posi- 
tive elongation,  or  from  any 
position  to  the  same  position  with 
motion  in  the  same  direction,  is 
given  by  the  equation 


FIG.  254. 


(10) 


Figure  2f54  represents  a  circle  of  radius  r.     The  angle  between 


310 


MECHANICS 


[ART.    185 


the  horizontal  diameter  and  the  radius  OB\s\\—t  radians.      The 


projection  of  OB  on  the  horizontal  diameter  is 


r  cos 


'Kg\ 


(11) 


If  the  point  B  moves  around  the  circle  with  uniform  speed, 
-  is  the  angle  which  the  line  OB  passes  through  in  one  second, 


and  if  t  is  the  time  which  has  elapsed  since  the  point  B  was  at  the 


right  end  of  the  horizontal  diameter,  then  \ —  t  gives  the  angle 

\   Tfl 

which  the  line  OB  makes  with  the  horizontal  diameter  at  the 
end  of  time  t.     If  tc  is  the  time  of  a  complete  revolution, 


tc  =  27r; 


27T 


m 
Kg' 


(12) 


Equation  (12)  shows  that  the  time  of  a  complete  revolution  of 

the  point  B  is  the  same 

H  F 

as  the  time  of  a  com- 
plete period  of  a  vibrat- 
ing mass. 

When  a  body  is  vi- 
brating under  the  action 
of  an  attractive  force 
which  is  proportional  to 
the  displacement,  its 
motion  along  its  path  is 
the  projection  on  that 
path  of  the  motion  of  a 
body  which  moves  with 
uniform  speed  around 
a  circular  path.  In  Fig. 
255,  A,  B,  C,  D,  E,  etc.  are  equidistant  points  on  the  circum- 
ference of  a  circle.  If  a  body  moves  in  this  circle  with  uniform 
speed,  it  passes  from  one  of  these  points  to  the  next  one  in  the 
same  interval  of  time.  The  points  B',  C',  D',  etc.  are  the  pro- 
jections of  the  points  B,  C,  D,  etc.,  on  the  horizontal  diameter  of 
the  circle,  The  body  which  is  vibrating  along  the  straight  line 


FIG.  255. 


CHAP.  XIX]     FORCE  VARIES  AS  DISPLACEMENT          311 

AM  passes  from  A  to  B>  ',  from  B'  to  C",  from  C"  to  Z)',  etc.,  in 
equal  intervals  of  time. 

A  motion  of  this  kind  is  called  a  simple  harmonic  motion.  A 
simple  harmonic  motion  is  equivalent  to  the  motion  in  the  diam- 
eter of  a  circle  of  the  projections  of  the  positions  of  a  body  which 
moves  with  uniform  speed  in  the  circumference  of  that  circle. 
The  circle  of  Fig.  255  may  be  called  the  reference  circle  of  the 
simple  harmonic  motion.  The  actual  motion  is  along  the  straight 
line  A  G'  M.  The  reference  circle  is  used  as  a  convenient  device 
for  finding  the  position  and  velocity  at  any  given  time. 

If  Equation  (7)  is  differentiated  with  respect  to  time,  the 
results  are 


dt 


md*x_ma_  • 

1  W~    7" 

Equation  (15)  is  identical  with  Equation  (2)  of  Art.  183.     The 
differentiation,  therefore,  checks  the  integration. 

The  velocity  of  the  simple  harmonic  motion  is  a  maximum 

when  the  sine  of  *  /  —  -  t  is  equal  to  unity.     From  Equation  (13), 


m 

the  maximum  velocity  is  r  A/— •     This  is  equal  to  the  velocity 

\  m 

of  a  body  which  is  moving  in  the  circumference  of  a  circle  of 

radius  r  and  passing  through  the  angle  of  A/ — -  radians    per 

\  m 

second. 

From  Fig.  255,  when  the  body  which  is  moving  in  the  reference 
circle  AG'M  is  passing  the  point  G',  its  motion  is  parallel  to  the 
diameter  AM.  The  projection  of  its  velocity  on  the  diameter  is, 
therefore,  equal  to  the  velocity  of  the  body  which  is  moving  in 
the  circle  of  reference. 

To  find  the  position  of  a  body  which  is  moving  with  a  simple 
harmonic  motion,  the  maximum  velocity  is  first  calculated  by 
means  of  the  energy  equation,  Formula  XXVI:  A  body  is 
assumed  to  be  moving  with  this  velocity  in  the  circumference 
of  a  reference  circle.  The  radius  of  this  reference  circle  is  equal 


312  MECHANICS  [ART.  185 

to  the  amplitude  of  the  vibration.  The  time  of  one  revolution 
of  the  body  in  the  reference  circle  gives  the  time  of  a  complete 
period  of  the  vibration.  The  position  at  any  time  is  found  by 
the  projection  on  the  diameter  of  the  reference  circle. 

Example 

Find  the  time  of  a  complete  vibration  of  the  body  of  Problem  2  of  Art.  184. 
Find  the  position  and  velocity  0.2  second  after  the  body  is  released. 

The  amplitude  is  3  ft.  and  the  circumference  of  the  circle  of  reference  is 
18.8946  ft.  The  maximum  velocity  was  found  to  be  15.53  ft.  per  sec. 
If  te  is  the  time  of  a  complete  period, 

18.8496 


At  0.2  second,  the  angle  traversed  in  the  reference  circle  is 

O  9 

Y~^  X  360  =  59.31°  =  59°  19'. 

re  =  3  cos  59°  19'  =  1.531  ft. 

v  =  15.53  sin  59°  19'  =  13.35  ft.  per  sec. 

The  time  of  a  complete  period  may  be  checked  by  Equation  (10) 
and  the  velocity  by  Equation  (13).  The  velocity  may  also  be 
checked  by  Formula  XXVI  after  the  displacement  has  been  found. 

It  is  recommended  that  the  student  make  use  of  the  reference 
circle  and  Formula  XXVI  instead  of  the  equations  of  this  article. 

The  potential  energy  of  a  stretched  spring  is  proportional  to 
the  square  of  the  displacement  from  the  position  of  equilibrium. 
The  kinetic  energy  is  proportional  to  the  square  of  the  velocity. 
The  maximum  velocity  is,  therefore,  directly  proportional  to  the 
amplitude.  The  circumference  of  the  reference  circle  is  propor- 
tional to  amplitude,  consequently  the  time  of  vibration  is  inde- 
pendent of  the  amplitude.  If  a  mass  on  a  given  spring  has  a 
maximum  velocity  of  1  foot  per  second  when  the  amplitude  of 
the  vibration  is  1  foot,  it  will  have  a  maximum  velocity  of  2  feet 
per  second  when  the  amplitude  of  the  vibration  is  2  feet.  In 
each  case  the  time  of  a  complete  period  is  2?r  seconds.  Equation 
(12),  which  gives  the  time  of  a  complete  period,  does  not  contain 
the  amplitude.  This  fact  is  a  mathematical  proof  of  the  state- 
ment that  the  time  of  vibration  is  independent  of  the  length  of 
the  swing.  A  vibration  which  is  independent  of  the  amplitude 
is  called  an  isochronous  vibration. 


CHAP.  XIX]     FORCE  VA  RIES  A  S  DISPLA  CEMENT          31 3 


Problems 

1.  The  mass  m  of  Fig.  253  is  40  pounds.     It  requires  a  pull  of  16  pounds 
to  displace  the  mass  a  distance  of  1  foot  in  the  direction  of  the  length  of 
the  spring.     The  mass  is  displaced  2  feet  from  the  position  of  equilibrium 
and  then  released.     Find  the  maximum  velocity.     Find  the  time  of  a 
complete  period.     Find  the  velocity  when  the  body  is  1  foot  from  the 
position  of  equilibrium. 

Ans.     Max  v  =  7.175  ft.  per  sec.;  te  =  1.75  sec. 

2.  In  Problem  1,  find  the  maximum  velocity  and  the  time  of  a  complete 
period  if  the  mass  is  displaced  1  foot  from  the  position  of  equilibrium  and 
then  released. 

3.  In  Problem  1,  the  mass  is  displaced  2  feet  and  then  released.     Find 
the  time  required  to  come  back  1  foot  toward  the  position  of  equilibrium. 


'*  Displacement 


FIG.  256. 

If  a  body  vibrating  with  a  simple  harmonic  motion  is  arranged 
to  draw  a  line  on  a  plate  which  moves  with  uniform  speed  at 
right  angles  to  the  direction  of  the  vibration,  the  line  so  drawn 
will  have  the  form  of  Fig.  256,  I. 

The  equation  for  the  velocity  of  a  simple  harmonic  motion  is 


v  =  — 


r  sm 


Kg\ 

ir  t  i 

m/ 


(13) 


314 


MECHANICS 


[ART.    186 


The  velocity  curve  is  a  sine  curve  which  has  the  same  period  as 
the  displacement  curve  of  Fig.  256,  I.  The  velocity  is  0  when 
the  displacement  is  a  maximum.  The  velocity  curve  is  shifted 
90  degrees  with  reference  to  the  displacement  curve,  as  is  shown 
in  Fig.  256,  II. 

To  an  observer  standing  in  front  of  an  engine,  the  circular 
motion  of  the  crank  pin  appears  as  a  linear  vertical  motion, 
which  is  rapid  at  the  middle  and  slower  at  the  top  and  the  bottom. 
This  is  a  simple  harmonic  motion,  if  the  engine  is  running  at 
uniform  speed.  To  an  observer  standing  at  one  side  of  the 
engine,  the  piston  rod  and  cross-head  appear  to  move  horizontally 
in  a  similar  way.  If  the  connecting  rod  were  infinitely  long, 


Connecting  Rod 


FIG.  257. 

this  would  be  a  simple  harmonic  motion.  With  a  connecting 
rod  of  finite  length,  as  shown  in  Fig.  257,  I,  the  motion  of  the 
cross-head  differs  slightly  from  a  true  simple  harmonic  motion. 

Figure  257,  II,  shows  a  method  of  changing  uniform  circular 
motion  into  horizontal  simple  harmonic  motion.  The  crank  B 
carries  a  pin  C  which  passes  through  a  vertical  slot  in  the  head  H. 
The  horizontal  component  of  the  circular  motion  is  transmitted 
to  the  horizontal  rod.  The  motion  of  the  rod  is,  therefore,  a 
true  simple  harmonic  motion,  when  the  angular  speed  of  the 
wheel  is  uniform. 

186.  Sudden  Loads. — When  a  load  is  applied  to  an  elastic 
body,  the  force  exerted  on  the  elastic  body  may  be  much  greater 
than  the  weight  of  the  load.  Figure  258  represents  a  helical 
spring  on  which  is  placed  a  mass  of  m  pounds.  Figure  258,  I, 
shows  the  natural  length  of  the  spring.  In  Fig.  258,  II,  the  mass 


CHAP.  XIX]     FORCE  VARIES  AS  DISPLACEMENT 


315 


m  is  attached  to  the  spring  but  is  held  up  by  the  support  B  so 
that  none  of  its  weight  comes  on  the  spring.  If  the  support  B 
is  lowered  slowly  until  all  the  weight  comes  on  the  spring,  the 
condition  of  Fig.  258,  III,  is  reached.  The  mass  comes  to  rest 


FIG.  258. 

when  the  spring  is  stretched  a  distance  d.     If  K  pounds  is  suffi- 
cient to  stretch  the  spring  1  foot, 

I  "  d  =  f  (1) 

The  work  done  by  gravity  on  the  mass  m  while  it  moves  a  dis- 
tance of  d  feet  is 

mi 

U~md"~  (2) 


316  MECHANICS  [ART.   186 

The  work  done  by  the  mass  on  the  spring,  or  the  potential  energy 
acquired  by  the  spring,  is 

TT  -  °  +  m  *  _  rad  _  ™2  /ON 

2  2    :"  2K 

Equation  (3)  represents  only  half  the  work  of  Equation  (2).  It 
is  necessary  to  account  for  the  remainder.  At  the  beginning  of 
the  descent,  the  entire  weight  of  the  body  rested  on  the  support 
B.  As  the  body  was  lowered,  the  spring  took  an  increasingly 
greater  proportion  until  it  finally  carried  the  entire  load  and  the 
reaction  of  the  support  became  zero.  The  work  done  on  the 
support  B  was 


Equation  (4)  accounts  for  the  remaining  half  of  the  energy. 
If  the  support  B  is  suddenly  removed  when  there  is  no  tension 
in  the  spring,  the  motion  is  that  shown  in  Fig.  258,  IV.  No 
work  is  done  on  the  support  and  all  the  energy  is  stored  in  the 
spring.  If  di  is  the  total  elongation  of  the  spring,  the  average 

tension  is  —  ~-  and  the  potential  energy  of  the  spring  is  -~^- 

Zi  2 

The  work  of  gravity  on  the  mass  is  mdi.  Equating  the  work  and 
energy, 

,        Kdl   . 
mdi  =  -3^;  (5) 

*  -    5'  .  v        (6) 

A  comparison  with  Equation  (1)  shows  that  the  elongation  when 
the  load  is  "  suddenly  applied"  is  twice  as  great  as  when  the 
load  is  gradually  applied.  When  a  load  is  suddenly  applied  to  a 
spring  or  other  elastic  body  in  which  the  stress  is  proportional 
to  the  deformation,  the  deformation  is  twice  as  great  as  that  due 
to  the  weight  of  the  load.  Since  the  force  is  proportional  to  the 
deformation,  it  follows  that  the  force  is  likewise  twice  as  great 
as  the  weight  of  the  load.  This  fact  is  expressed  by  the  state- 
ment "Live  load  is  twice  the  dead  load." 

Since  the  tension  in  the  spring  at  the  bottom  of  its  path  in 
Fig.  258  is  twice  the  weight  of  the  body,  the  resultant  force 
upward  in  that  'position  is  m  pounds.  The  body  will  move  up- 
ward to  the  position  of  Fig.  258,  II,  and  will  continue  to  vibrate 


CHAP.  XIX]     FORCE  VARIES  AS  DISPLACEMENT          317 

for  a  long  time.     It  will  finally  come  to  rest  in  the  position  of  Fig. 
258,  III. 

In  Fig.  258,  V,  the  body  is  suspended  from  the  spring  by  an 
inelastic  cord  which  permits  it  to  fall  a  distance  h  before  it  brings 
any  tension  on  the  spring.  If  dz  is  the  distance  which  the  spring 
is  stretched,  the  work  of  gravity  on  the  mass  is  m(h  +  d2),  and 

Kd* 
the  work  done  on  the  spring  is  —    - 


m(h  +  d.)  -          -  (7) 

When  a  load  falls  upon  an  elastic  body,  the  deformation  is  greater 
than  twice  the  deformation  caused  by  the  weight  of  the  load. 
The  body  will  vibrate  according  to  the  laws  of  a  simple  harmonic 
motion  from  the  position  of  no  deformation  of  the  elastic  body 
down  to  the  lowest  point  and  back  again.  After  it  reaches  the 
position  at  which  there  is  no  deformation  of  the  elastic  body,  the 
load  will  continue  to  move  as  a  projectile  which  is  thrown  straight 
upward.  After  reaching  the  starting  point,  it  will  fall  and  repeat 
the  cycle. 

Problems 

1.  A  force  of  10  pounds  stretches  a  given  spring  6  inches.     How  much  is 
this  spring  stretched  when  a  mass  of  40  pounds  is  hung  on  it  so  that  the  load 
is  gradually  applied? 

2.  How  much    is  the    spring   of   Problem    1 
stretched  if  the  mass  of  40  pounds  is  put  on  sud- 
denly?    What  is  the  tension  at  the  lowest  point? 
What  is  the  time  required  to  reach  the  lowest 
point? 

3.  The  40-pound  mass  of  Problem  1  falls  on 
the  spring  from  a  height  of  1  foot.     How    much 
is  the  spring  elongated?  Ans.     4.828  ft. 

4.  In    Problem    3,    what   is   the   entire    time 

required    for    the    mass    to    fall,   compress  the  j,      259 

spring,  and  return  to  its  original  position? 

At  the  position  of  equilibrium,  the  spring  is  stretched  2  feet.  The 
amplitude  of  the  vibration  below  this  point  is  2.828  feet.  Consequently, 
the  radius  of  the  circle  of  reference  is  2.828  feet.  The  angle  6  of  Fig.  259 
is  given  by  the  equation 


COS 


If  the  entire  vibration  were  a  simple  harmonic  motion,  the  time  of  the  com- 
plete period  would  be  tc  =  1.566  seconds.     The  time  required  to  traverse 


318  MECHANICS  [ART.  187 

270  degrees  of  the  reference  circle  is  1.174  seconds.     The  time  required  to 
fall  1  foot  is  0.249  second.     The  entire  time  is  1.174  +  0.498  =  1.672  sec. 

187.  Composition  of  Simple  Harmonic  Motions. — Figure  260 
shows  a  body  supported  by  a  single  vertical  spring  and  held  by 
two  horizontal  springs.  [One  horizontal  spring  is  sufficient  if  it 
does  not  buckle  in  compression.]  The  springs  are  supposed  to 
be  so  long  that  the  vertical  component  of  the  tension  in  the  hori- 
zontal springs  when  the  body  is  displaced  vertically  may  be 
neglected,  and  the  horizontal  component  in  the  vertical  spring 
when  the  body  is  displaced  horizontally  may  be  also  neglected. 
If  the  body  is  displaced  horizontally,  it  will  vibrate  in  that  direc- 
tion with  a  simple  harmonic  motion.  If  it  is  displaced  vertically, 


FIG.  260. 

it  will  vibrate  vertically  with  a  simple  harmonic  motion.  If  it  is 
displaced  to  the  right  a  distance  a  and  given  a  vertical  component 
upward  sufficient  to  carry  it  to  a  vertical  distance  6,  it  will  then 
have  a  motion  which  is  the  resultant  of  the  two  simple  harmonic 
motions  at  right  angles  to  each  other.  If  Kh  is  the  constant  of 
the  horizontal  springs,  and  Kv  is  the  constant  of  the  vertical 
spring,  the  equations  of  displacement  are 


a  cos  .  ^  t,  CD 


2,  =  6  sin  ,  f-^  «.  (2) 


If  Kh  =  KV)  the  periods  of  the  two  components  are  equal  and 
the  path  of  the  body  is  an  ellipse.  If  the  amplitudes  are  equal, 
the  ellipse  becomes  a  circle. 


CHAP.  XIX]     FORCE  VARIES  AS  DISPLACEMENT 


319 


Figure  261  shows  the  motion  when  the  time  of  vibration  in  the 
horizontal  direction  is  twice  as  great  as  in  the  vertical  direction. 
The  equations  may  be  written  briefly, 

x  =  a  sin  ct;  y  =  b  sin  2ct. 


FIG.  261. 


Problems 

1.  If  the  force  required  to  stretch  the  horizontal  spring  1  foot  is  20 
pounds,  what  must  be  the  force  required  to  stretch  the  vertical  spring  1 
foot  in  order  that  the  time  of  vibration  vertically  may  be  one-half  the  time 
of  vibration  horizontally? 

2.  Plot  the  curve  of  the  resultant  of  two  simple  harmonic  motions  of 
equal  amplitude  if  the  time  of  vibration  horizontally  is  two-thirds  the  time 
of  vibration  vertically. 

Instead  of  the  springs  as  shown  in  Fig.  260,  a  single  bar  may 
be  used  as  the  elastic  body.  The  bar  is  fixed  rigidly  at  one  end 
and  carries  a  heavy  mass  at  the  other  end.  The  section  of  the 


320  MECHANICS  [ART.   188 

bar  is  a  rectangle.  In  Fig.  262,  the  vertical  sides  of  the  rectangle 
are  greater  than  the  horizontal  width.  The  vertical  stiffness  is 
greater  than  the  horizontal  stiffness  and  the  time  of  vibration  in 
the  vertical  plane  is  greater  than  in  the  horizontal  plane.  These 
dimensions  may  be  such  as  to  give  any  desired  ratio  to  the  time  of 
vibration. 


Front  View 
FIG.  262. 

Problem 

(A  knowledge  of  Strength  of  Materials  is  required  for  this  problem.) 
3.  A  steel  cantilever  30  inches  long  is  Y±  inch  wide  and  %  inch  deep.  It 
carries  a  mass  of  2  pounds  on  the  free  end.  If  E  =  30,000,000,  find  the 
force  required  to  deflect  this  cantilever  1  inch  in  a  vertical  direction  and  the 
force  required  to  deflect  it  1  inch  in  the  horizontal  direction.  Calculate 
Kv  and  Kh.  Neglecting  the  mass  of  the  bar,  find  the  time  of  a  complete 
period  in  each  direction. 

188.  Correction  for  the  Mass  of  the  Spring. — In  the  case  of  a 
vibrating  spring  or  other  elastic  body,  a  part  of  the  kinetic  energy 
is  located  in  the  spring.  For  accurate  work,  correction  must  be 
made  for  this  energy. 

The  effective  mass  of  the  spring  may  be  calculated  from  two 
sets  of  vibration  experiments.  Let  m0  be  the  effective  mass  of 
the  spring,  mi  be  a  known  additional  mass,  and  ra2  be  a  second 
known  mass.  The  time  of  vibration  is  found  with  the  mass  mi 
on  the  spring,  and  again  with  mz  on  the  spring.  If  ti  and  t2  are 
these  times  of  vibration, 

CD 


Kg 


,,  =  fc^SJjSSj  (2) 

i  =  2r+-S;  ;          (3) 

/wi        -1—  -f  2 

ra2  -  mi  =  t\  -  t\  (*\ 


CHAP.  XIX]     FORCE  VARIES  AS  DISPLACEMENT          321 


Problem 

The. time  of  vibration  of  a  given  spring  with  a  load  of  2  pounds  is  1.640 
seconds.  The  time  of  vibration  with  a  load  of  10  pounds  is  2.880  seconds. 
Find  the  effective  mass  of  the  spring. 

Ans.     m0  =  1.839  Ib. 

The  effective  mass  of  a  helical  spring  may  easily  be 
calculated.  Let  m  be  the  mass  of  a  helical  spring 
of  length  I,  as  shown  in  Fig.  263.  The  mass  per 

tn 
unit  length  is  y     The  mass  of  an  element  of  length 

7??  f7  77 

dy  is  —j —  If  v  is  the  velocity  of  the  free  end  of  the 
spring  at  a  distance  I  from  the  fixed  end,  the  velocity 
of  an  element  at  a  distance  y  from  the  fixed  end  is  y 
The  energy  of  the  element  is  given  by  the  equation 


dU  = 


£7=^1^ 


U 


mv 


Equation  (7)  shows  that  the  effective  mass  of  a  uniform  helical 
spring  is  one-third  the  mass  of  the  spring.  This  statement 
applies  to  the  helical  part  of  the  spring.  Where  the  spring 
bends  toward  the  axis  at  the  ends,  the  ratio  is  different.  In  a 
long  spring  with  a  large  number  of  turns,  the  error  which  is  due 
to  the  end  connections  may  be  neglected. 

189.  Determination  of  g. — The  vibration  of  a  body  on  a  helical 
spring  may  be  used  for  the  determination  of  g.  The  time  of 
vibration  is  determined  with  a  known  mass  on  the  spring;  the- 
elongation  of  the  spring  under  a  known  load  is  measured;  and 
the  spring  is  weighed.  The  results  of  these  measurements  afford 
the  necessary  data  for  the  calculation  of  gravity. 

;"••  EXAMPLE 

A  spring  which  weighs  0.24  pound  is  elongated  0.421  ft.  when  a  load  of 
5  pounds  is  applied.    With  the  5-pound  mass  on  the  spring,  the  time  of 
21 


322  MECHANICS  [ART.  190 

100    complete    periods   is   72.4  seconds.    Find  the  value   of  g  for  that 
locality. 


-;  m  =  5  +  0.08;  tc  =  0.724  sec. 


0.4: 

47r2  X  5.08  X  0.421 
5  X  0.7242 


=  32.20  ft. 


This  experiment  does  not  require  that  the  masses  be  accurately 
known  in  terms  of  the  standard  units.  If  the  effective  mass  of 
the  spring  were  negligible,  it  would  only  be  necessary  to  find  the 
elongation  due  to  any  given  mass  and  the  time  of  vibration  with 
that  mass.  Since  the  effective  mass  of  the  spring  is  not  negli- 
gible, the  ratio  of  its  mass  to  that  of  the  additional  load  must  be 
determined.  It  is  not  necessary,  however,  to  determine  the 
actual  mass  of  either  the  spring  or  the  load. 

Problem 

A  helical  spring  weighing  0.42  pound  is  stretched  0.362  foot  by  a  load  of 
6  pounds.  With  a  mass  of  10  pounds  on  the  spring,  the  time  of  100  com- 
plete periods  is  86.6  seconds.  Find  g. 

190.  Determination  of  an  Absolute  Unit  of  Force. — If  an 

absolute  unit  of  force  is  used,  Equations  (2)  and  (3)  of  Art.  184 
become 

-K'x  =  ma;  (1) 

-X'*  =  mg-  .          ^         (2) 

In  these  equations,  K'  is  the  force  in  absolute  units  which  causes 
unit  displacement. 
Formula  XXVI  is 

mv2  _  K'r2  _  K'x2 

~2~       ~2~~       ~T~ 

Equation  (12)  of  Art.  185  is 

te  =  2^~  (4) 

Equation  (4)  affords  an  easy  method  of  finding  the  constant  of 
the  spring  and  getting  a  unit  of  force  which  is  entirely  inde- 
pendent of  gravity. 


CHAP.  XIX]     FORCE  VARIES  AS  DISPLACEMENT          323 

Problems 

1.  A  helical  spring  weighs  0.6312  pound.     A  10-pound  "weight"  is  hung 
on  this  spring  and  the  time  of  1000  complete  periods  is  found  to  be  1243.6 
seconds.     Find  the  constant  of  the  spring  in  poundals  if  the  actual  mass  of 
the  10-pound  weight  is  9.9842  pounds. 

Ans.     K'  =  260.2  poundals;  100  poundals  elongates  the  spring  0.3844  ft. 

2.  A  given  spring  makes  1000  complete  vibrations  in  528.3  seconds  when 
a  500  gram  mass  is  attached  to  it,  and  makes  1000  complete  vibrations  in 
1014.2  seconds  when  the  mass  is  increased  to  2000  grams.     How  many 
dynes  force  are  required  to  deform  this  spring  1  centimeter? 

Ans.     K'   =  7900.9  dynes. 

191.  Positive  Force  which  Varies  as  the  Displacement. — The 

reaction  of  a  spring  is  proportional  to  the  displacement.  The 
direction  of  the  reaction  is  opposite  the  direction  of  the  displace- 
ment. The  expression  for  the  reaction  of  the  spring  is  P  =  —Kx. 
When  the  force  varies  as  the  displacement  and  is  in  the  same 
direction,  the  expression  is 

P  =  Kx.  (1) 

If  a  body  subjected  to  a  force  of  this  kind  is  in  equilibrium, 
the  equilibrium  is  unstable.  If  the  body  is  displaced  ever  so 
little  from  its  position  of  equilibrium,  the  increased  force  in 
the  direction  of  the  displacement  will  cause  it  to  continue  to 
move  in  the  same  direction,  and  it  will  not  return  to  the  position 
of  equilibrium.  If  the  body  is  displaced  a  distance  r  from  the 

Kr2 
position  of  equilibrium,   the  work  done  by  the  force  is  -5-. 

The  work  done  in  displacing  the  body  from  a  distance  r  to  a 
distance  x  from  the  position  of  equilibrium  is 

U  =  £(*>  -  r«).  (2) 

If  the  body  is  at  rest  when  the  displacement  is  r,  its  kinetic 
energy  when  its  displacement  is  x  is  given  by  the  equation 

mv2       K,  „         2,  ,QN 

-w " :  ~2(x  ' r}- 

Problems 

1.  A  rope,  100  feet  long,  weighing  1  pound  per  foot,  runs  over  a  smooth 
pulley  as  shown  in  Fig.  264.  When  the  center  of  the  rope  is  1  foot  below 
the  pulley,  so  that  there  is  49  feet  of  rope  on  one  side  of  the  pulley  and  51 
feet  on  the  other  side,  what  is  the  resultant  force?  Ans:  P  =  2  Ib. 


324 


MECHANICS 


[ART.    192 


2.  In  Problem  1,  what  is  the  expression  for  the  resultant  force  in  terms  of 
the  displacement  of  the  center  of  the  rope?  Ans.     P  =  2x. 

3.  Find  the  work  done  when  the  rope  of  Fig.  264  moves  from  the  position 
of  equilibrium  to  the  position  at  which  there  is  40  feet  of  rope  on  one  side 
and  60  feet  on  the  other.  Ans.     U  =  100  ft.-lb. 

4.  What  is  the  final  velocity  in  Problem  3  if  the  initial  velocity  is  so  small 
that  its  kinetic  energy  is  negligible?  Ans.     v  =  8.02  ft.  per  sec. 


The  relation  of  displacement  to  time  is  found  by 
integration.     From  Equation  (3) 


(4) 
(5) 
(6) 


If     time     is     measured    from     the    time    when    x 
=    r,     then    t   =   0   when   x  =    r.     Substituting     in 


FIG.  264.    Equation  (6), 


(7) 
(8) 

(9) 


Problem 


6.  In  Fig.  264,  the  rope  is  placed  with  49  feet  on  one  side  and  51  feet  on 
the  other  and  then  released.  Find  the  time  required  for  the  rope  to  entirely 
leave  the  pulley.  Find  the  final  velocity. 

Ans.     t  =  5.63  sec.;  v  =  40.10  ft.  per  sec. 

192.  Summary. — There  are  two  classes  of  forces  which  vary 
as  the  displacement.  In  one  class,  the  direction  of  the  force  is 
opposite  the  direction  of  the  displacement.  In  the  other  class, 
the  direction  of  the  force  is  the  same  as  the  direction  of  the 
displacement. 

A  spring  or  other  elastic  body  exerts  a  force  which  is  opposite 
the  displacement.  The  expression  for  the  force  is 

P  =  -Kx. 


CHAP.  XIX]     FORCE  VARIES  AS  DISPLACEMENT          325 

The  change  of  kinetic  energy  when  the  body  moves  from  displace- 

Kr2       Kx2 
ment  r  to   displacement   x  is  —^- ^—     If  the  velocity  is 

zero  at  displacement  r,  the  velocity  at  displacement  x  is  given  by 
the  energy  equation, 

mv2      Kr2       Kx2  _         .    ____„_ 

-£—  =  -g 2~  Formula  XXVI 

When  the  force  is  proportional  to  the  displacement  and  in  the 
opposite  direction,  the  motion  is  periodic.  The  position  at  any 
time  is  given  by  the  equation 

x  =  r  cos 

in  which  r  is  the  maximum  displacement,  or  the  amplitude  of  the 
vibration,  t  is  the  time  which  has  elapsed  since  the  displace- 
ment was  equal  to  r,  and  x  is  the  displacement  at  time  t.  The 
velocity  at  time  t  is  given  by  the  equation 

dx 

~  dt 

The  time  of  a  complete  period  is 


The  time  of  vibration  is  independent  of  the  amplitude. 

Simple  harmonic  motion  may  be  studied  by  means  of  a  refer- 
ence circle,  the  radius  of  which  is  equal  to  the  amplitude  of 
the  motion.  The  maximum  velocity  is  found  by  Formula 
XXVI.  A  body  is  assumed  to  move  with  this  velocity  in  the 
circumference  of  the  reference  circle.  The  projection  of  the 
motion  on  the  diameter  of  the  reference  circle  gives  the  required 
motion. 

When  a  load  is  applied  suddenly  to  an  elastic  body,  the 
deformation  is  twice  as  great  as  that  when  the  load  is  applied 
gradually,  and  the  stress  is  twice  as  great  as  that  resulting  from 
a  quiescent  load. 

An  absolute  unit  of  force  may  be  determined  by  means  of  the 
time  of  vibration  of  a  known  mass  on  a  spring.  The  value  of 
g  may  be  found  by  means  of  the  time  of  vibration  of  a  mass  on 
a  spring  and  the  elongation  of  the  spring  caused  by  the  weight 
of  the  mass. 


326  MECHANICS  [ART.  192 

The  second  class  of  force  which  is  proportional  to  the  displace- 
ment is  expressed  by  the  equation 

P  =  Kx. 

If  the  velocity  is  zero  at  displacement  r,  the  velocity  at  dis- 
placement x  is  given  by  the  energy  equation 

mv*      Kx2       Kr2 
2g    "       2  2  ' 

The  motion  is  not  periodic.  The  equilibrium  when  x  =  0  is 
unstable.  Equations  (8)  and  (9)  of  Art.  191  give  the  position 
in  terms  of  the  time. 


CHAPTER  XX 
CENTRAL  FORCES 

193.  Definition. — A  central  force  is  one  whose  direction  is  along 
a  line  which  joins  its  point  of  application  wtih  some  fixed  point; 
whose  magnitude  depends  upon  the  length  of  that  line  and  does 
not  depend  upon  its  direction.  In  Fig.  265,  0  is  a  central  point 


FIG.  265. 


FIQ.  266. 


and  Q  is  the  point  of  application  of  a  force  directed  toward  0. 
If  the  force  satisfies  the  definition  of  a  central  force,  its  magnitude 
is  constant  for  any  position  of  Q  on  the  surface  of  a  sphere  of 
radius  OQ  with  its  center  at  0. 


FIG.  267. 

In  Fig.  266,  a  rope  which  supports  a  load  W  runs  over  a  small 
pulley.  The  force  at  B  is  constant  and  is  directed  toward  the 
pulley  as  a  center. 

In  Fig.  267,  the  rope  which  runs  over  the  pulley  is  attached  to  a 
spring.  If  the  end  of  the  rope  is  at  a  distance  b  from  the  pulley 

327 


328  MECHANICS  [ART.   194 

when  there  is  no  tension  in  the  spring,  the  force  at  B  is  given  by 
the  equation, 

P  «  -  K(r  -  V).  (1) 

The  force  of  gravity  is  a  central  force  directed  toward  the 
center  of  the  Earth.  Above  the  surface  of  the  Earth,  the  magni- 
tude of  the  force  of  gravity  varies  inversely  as  the  square  of  the 
distance  from  the  center.  The  force  is  given  by  the  expression, 


An  expression  similar  to  Equation  (2)  gives  the  attraction  or 
repulsion  of  a  single  magnetic  pole  or  of  an  electrically  charged 
sphere. 

194.  Work  of  a  Central  Force.  —  The  work  done  by  a  central 
force  depends  upon  the  distances  of  the  initial  and  final  points  of 
application  from  the  center,  and  is  independent  of  the  form  of 
the  path. 


Fio.  268.  FIG.  269. 

In  Fig.  268,  the  point  of  application  moves  a  distance  ds  from 
point  A  to  point  B.  The  angle  between  the  radius  vector  OA 
and  the  line  AB  is  <J>.  The  work  of  the  displacement  is 

dU  =  P  cos  <f>  ds  =  Pds  cos  <}>.  (1) 
Since  ds  cos  <t>  =  dr, 

dU  =  Pdr;  (2) 

U  =  fPdr.  (3) 

If  a  body  under  the  action  of  a  force  which  is  central  at  0,  Fig. 
269,  moves  from  a  point  E  at  a  distance  r\  from  0  to  a  point  F 
at  a  distance  r2  from  0,  the  total  work  depends  only  upon  the 
distance  r2  —  r\.  The  body  might  move  from  E  to  G  on  the  cir- 
cumference of  a  circle  of  radius  r\.  Since  the  force  is  normal  to 
the  displacement,  no  work  is  done  during  this  displacement.  The 


CHAP.  XX] 


CENTRAL  FORCES 


329 


body  may  then  move  along  the  line  OGF  directly  away  from  the 
center.     The  total  work  is  equal  to  that  given  by  Equation  (3). 

Problems 


- 


1.  A  body  rests  on  a  horizontal  plane.     It  is  subjected  to  a  constant  force 
of    20   pounds    directed  toward 

a  point  0  which  is  12  feet 
above  the  plane,  and  directly 
over  a  point  B  on  the  plane. 
The  body  moves  from  a  point 
which  is  16  feet  from  B  to  a 
point  which  is  5  feet  from  B. 
Find  the  work  done  by  the 
force  of  20  pounds.  Solve  by 
arithmetic. 

2.  Solve  Problem  1  by  integra- 
tion,   assuming    that    the    body      %w .%xw .WN%WWWW 

moves  along  a  straight  line  on            £'_"_"  ^'"""^       x       J 
the  surface  of  the  plane.  f .* /$'— 

From  Fig.  270, 


*s{/////jtyf 

lf\ 


13' 


JL_L±:. 


\ 

> 


? 

.:£_ 


FIG.  270. 

dU  =  20  cos  6  dx  -  -    2°X&^. 
U  =  140  ft.-lb. 

3.  The  body  of  Problem  1  weighs  60  pounds.  If  its  velocity  is  zero  when 
it  is  16  feet  from  B,  what  is  its  velocity  when  it  is  5  feet  from  B?  If  its 

velocity  is  20  feet  per  second 
when  it  is  16  feet  from  B, 
what  is  its  velocity  when  it 
is5feetfrom£? 
Ans.  v  =  12.26  ft.  per  sec.; 
v  =  23.45  ft.  per  sec. 
4.  The  body  of  Problem  1 
weighs  60  pounds  and  the 
coefficient  of  friction  be- 
tween the  body  and  the 
plane  is  0.1.  Find  the  in- 
crease of  kinetic  energy  when 
the  body  moves  from  the 
point  at  16  feet  from  B  to 
the  point  at  5  feet  from  B. 
Find  the  friction  by  integ- 
ration. 

Ans.     94.63  ft.-lb. 

6.  Figure  271  shows  a  drum  which  is  rotated  by  a  spiral  spring.  A  rope 
attached  to  the  drum  runs  over  a  smooth  pulley  0  at  a  distance  of  20  feet 
above  a  point  B  on  a  smooth  horizontal  floor.  There  is  no  tension  in  the 
rope  when  the  end  is  at  C  at  a  distance  of  5  feet  from  0.  A  force  of  3  pounds 


330  MECHANICS  [ART.  195 

is  required  to  rotate  the  drum  1  foot.  The  drum  is  rotated  and  the  end  of 
the  rope  is  fastened  to  a  mass  of  80  pounds  which  rests  on  the  floor.  The 
80-pound  mass  moves  from  a  point  which  is  21  feet  from  B  to  a  point  which 
is  15  feet  from  B.  Find  the  work  done  by  the  spring.  If  the  mass  of  the 
drum  and  spring  is  negligible  and  the  80-pound  mass  starts  from  rest,  find 
its  final  velocity.  Ara.  U  =  264  ft.-lb. 

6.  Solve   Problem  5  if  the  coefficient  of  fric- 
tion between  the  mass  and  the  floor  is  0.12. 

Ans,     Change  of  kinetic  energy  =  2  64  — 
57.6  +  43.2  -  8.03  =  241.57  ft.-lb. 

7.  Figure  272  is  similar  to  Fig.  270.    The  force, 
however,  is  that  resulting  from  a  suspended  mass 
of  20  pounds.     The  body  starts  from  a  point  16 
feet   from   B.     If   there   is  no  friction,  find  its 
velocity  when  it  reaches  B. 

Ans.    v  =  13.1  ft.  per  sec. 

8.  In   Fig.    272,  find  the  velocity  of  the  60- 
FIG  272.                   pound  mass  at  5  feet  from  B  if  its  velocity  was 

zero  at  16  feet  from  B.     Find  also  the  velocity 
of  the  20-pound  mass. 

Ans.     Velocity  of  the  60-pound  mass  =  11.91  ft.  per  sec. 

195.  Force  Inversely  as  the  Square  of  the  Distance.  —  When 
the  magnitude  of  the  force  varies  inversely  as  the  square  of  the 
distance  from  the  central  point  and  is  directed  toward  that  point, 
its  equation  is 


When  the  point  of  application  is  displaced  a  distance  dr,  the 
increment  of  work  done  by  the  force  is 

-  -      (2) 


If  the  total  displacement  is  r2  —  ri,  the  total  work  is 

-i 

r 

Problems 

1.  The  attraction  between  two  charged  spheres  is  200  dynes  when  their 
centers  are  5  centimeters  apart.     Find  K  of  the  equations  above.     How 
much  work  is  done  in  moving  one  sphere  from  a  position  10  centimeters  from 
the  other  to  a  position  20  centimeters  from  the*  other? 

Ans.     K  =  5000  dynes;   U  =  500  ergs. 

2.  For  bodies  above  the  Earth's  surface,  the  attraction  of  gravitation 
varies  inversely  as  the  square  of  the  distance  from  the  center  of  the  Earth. 
If  the  radius  of  the  Earth  is  taken  as  4000  miles,  and  is  expressed  in  feet, 
what  is  the  value  of  K  in  pounds?  Ans.     K  =  21,120,0002. 


CHAP.  XX]  CENTRAL  FORCES  331 

3.  A  mass  of  1  pound  is  lifted  from  the  surface  of  the  Earth  a  vertical 
distance  of  4000  miles.     Find  the  work  in  foot-pounds. 

Ans.     U  =  10,560,000  ft.-lb. 

4.  If  only  the  attraction  of  the  Earth  is  considered,  show  that  the  work 
done  in  moving  a  body  from  the  surface  of  the  Earth  to  an  infinite  distance 
is  21,120,000  foot-pounds  multiplied  by  the  mass  of  the  body. 

5.  A  body  falls  to  the  Earth  from  an  infinite  distance.     If  the  resistance 
of  the  air  were  negligible,  with  what  velocity  would  it  strike  the  Earth? 

Ans.     37,000  ft.  per  sec. 

If  a  body  were  thrown  out  of  the  Earth's  atmosphere  with 
a  velocity  of  37,000  feet  per  second,  it  would  not  return.  It 
would  not,  however,  leave  the  solar  system  on  account  of  the 
greater  energy  required  to  overcome  the  attraction  of  the  Sun. 

The  results  of  Problems  2,  3,  and  4  are  slightly  in  error  owing 
to  the  fact  that  4000  miles  has  been  used  as  the  radius  of  the 
Earth.  The  actual  radius  is  less  than  21,000,000  feet.  (See 
Hussey's  Mathematical  Tables,  p.  147.) 

196.  Equipotential  Surfaces. — In  Art.  135,  an  equipotential 
surface  was  denned  as  a  surface  normal  to  the  resultant  of  all 
the  applied  forces  except  the  reaction  of  the  surface.  No  work 
is  done  when  a  body  moves  from  one  part  of  an  equipotential 
surface  to  another.  When  the  forces  have  a  single  center,  the 
equipotential  surfaces  are  concentric  spheres. 

If  U  is  the  potential  energy  of  a  body,  and  if  the  body  receives 
a  displacement  dx  while  y  and  z  remain  constant,  the  change 
in  the  energy  is  the  partial  derivative  of  U  with  respect  to  x 
multiplied  by  dx.  When  x  alone  varies, 

increment  of  U  '=  —  dx.  (1) 

oX 

Since  the  work  done  in  moving  a  distance  dx  is  the  product  of 
dx  multiplied  by  the  component  of  the  force  in  that  direction, 
the  partial  derivative  of  U  with  respect  to  X  is  the  X  component 
of  the  applied  forces. 

dU      „     dU  dU  __  (2. 

^x=Hx'^  =    V'~dz 

The  equation  of  the  equipotential  surface  is 

dU  ,      .    dU  ,      ,dU,  f^ 

~dx  +  —dy  +  —dz==0. 

The  potential  energy  U  is  called  the  potential  function.  The 
potential  function  may  be  denned  as  the  expression  whose 


332 


MECHANICS 


[ART.    196 


derivative  with  respect  to  any  direction  gives  the  force  in  that 
direction. 

Problems 

1.  Solve  Problem  1  of  Art.  133  by  Equation  3.     Assume  that  the  diameter 
of  the  pulley  is  negligible. 

If  the  origin  of  coordinates  is  taken  at  the  pulley,  Fig.  273,  the  energy 

p2  /J.2         I        y2 

due  to  gravity  is  Wy.     The  energy  of  the  spring  is  =-  = •= — 


U  =  Wy 


dU 


dU 


2 
10 


dx   ~     ''  dy 
xdx  +  (10  +  y)dy  =  0; 


FIG.  273. 


FIG.  274. 


2.  A  body  is  subjected  to  the  action  of  two  forces  directed  toward  two 
fixed  points,  A  and  B  of  Fig.  274,  at  a  distance  a  apart.     The  magnitude  of 
each  force  is  Kr  in  which  r  is  the  distance  from  A  for  one  force  and  the 
distance  from  B  for  the  other.     Find  the  equation  of  the  equipotential 
curves  in  a  plane  through  A  and  B. 

If  B  is  taken  as  the  origin  of  coordinates, 

2U  =  K((a  +  xY  +  y*)  +  K(x*  +  y«); 

[2(a  +  z)  +  2x]dx  +  [2y  +  2y]dy  =  0; 

adx  +  2xdx  +  2ydy  =0; 

ax  +  x2  +  y2  =  C. 

The  curves  are  circles  with  their  centers  midway  between  A  and  B. 

3.  In  Fig.  274,  let  the  distance  from  A  to  B  be  10  feet.    Let  the  force 
which  is  central  at  A  be  3r,  where  r  is  the  distance  from  A ;  and  let  the  force 
which  is  central  at  B  be  2r,  where  r  is  the  distance  from  B.     With  the 
origin  of  coordinates  at  B  and  with  the  X  axis  passing  through  A,  find  the 
equation  of  the  equipotential  curve  through  the  point  (0,8). 

Ans.     (x  +  6)2  +  y2  =  100. 


CHAP.  XX]  CENTRAL  FORCES  333 

The  curve  is  a  circle  with  its  center  on  the  line  AB  at  a  distance  of  6  feet 
from  B. 

4.  In  Problem  3,  find  the  potential  energy  at  the  point  (0,8)  and  at  the 
point  on  the  X  axis  at  the  same  distance  from  the  center  of  the  circle 
Compare  the  two  values. 

197.  Summary.  —  The  direction  of  a  central  force  is  along  the 
line  which  joins  its  points  of  application  with  some  center;  its 
magnitude  is  some  function  of  the  length  of  this  line. 

The  work  done  by  a  central  force  depends  only  upon  the  dif- 
ference of  the  distances  of  the  initial  and  final  positions  from  the 
center. 

U  =  Pfdr 

When  the  magnitude  of  a  force  varies  inversely  as  the  square 
of  distance  of  its  point  of  application  from  the  center,  the  poten- 
tial difference  between  two  points  is 


r2 
If  r2  is  infinity, 


which  is  the  potential  of  the  point.  If  the  force  is  attraction,  the 
potential  is  the  work  done  by  the  force  in  bringing  the  body  from 
infinity  to  the  point  at  a  distance  r  from  the  center.  If  the 
force  is  repulsion,  the  potential  is  the  work  done  against  the 
force  in  bringing  the  body  from  infinity. 

The  potential  due  to  several  forces  is  found  by  adding  the 
potential  due  to  the  separate  forces.  Since  potential  is  not  a 
vector,  this  addition  is  the  addition  of  ordinary  arithmetic.  The 
derivative  of  the  potential  with  respect  to  any  direction  gives  the 
force  in  that  direction, 

dU       „     dU       T7  dU       „ 
to-*-;aT    V>lte" 

The  equation  of  an  equipotential  surface  is 

dU,        dU,        dU, 

dz  =  0 


CHAPTER  XXI 
ANGULAR    DISPLACEMENT    AND    VELOCITY 

198.  Angular    Displacement.  —  When    a    rigid    body    rotates 
about  an  axis,  all  parts  of  the  body  turn  through  the  same  angle. 

In  Fig.  275,  the  axis  passes  through  0  and  is 
perpendicular  to  the  plane  of  the  paper.  If 
a  point  on  the  body  moves  from  position  B 
to  position  B',  its  path  is  the  arc  BBf.  The 
angular  displacement  of  this  point  is  the 
angle  between  the  radii  OB  and  OB'.  If  r 
is  the  length  of  the  radius  from  0  to  B,  the 
FIG.  275.  angular  displacement  in  radians  is 

_  arc  BB' 

\J     —  * 

r 

Angular   displacement   is   linear   displacement   divided   by  the 
radius. 

Problems 

1.  A  point  moves  a  distance  of  6  feet  in  the  circumference  of  a  circle  of 
3-foot  radius.     Find  the  angular  displacement  in  radians. 

Ans.     0  =  2  radians. 

2.  A  body  revolves  once  on  its  axis.     What  is  the  angular  displacement? 

Ans.     6  =  2ir  =  6.2832  radians. 

3.  What  is  the  angular  displacement  when  a  body  turns  through  an  angle 
of  60  degrees? 

4.  A  cylinder  4  feet  in  diameter  turns  through  an  angle  of  50  degrees. 
Find  the  linear  displacement  of  any  point  on  the  surface  of  this  cylinder. 

5.  The  hour  hand  of  a  clock  moves  from  2  to  10.     What  is  the  angular 
displacement? 

199.  Angular  Velocity.  —  Angular  velocity  is  the  rate  of  change 
of  angular  displacement.     Angular  velocity  is  usually  represented 
by  the  Greek  letter  co  (omega). 

•- 


Equation  (1)  is  analogous  to  the  corresponding  formula  for  linear 
velocity.     For  small  intervals  of  time, 


334 


CHAP.  XXI]         ANGULAR  DISPLACEMENT  335 

If  v  is  the  linear  velocity  of  a  point  on  a  rotating  body,  vt 
is  the  length  of  the  arc  which  the  point  describes  in  time  t.     The 

angular  displacement  of  the  point  is  —  The  angular  velocity  of 
the  point  is 

?*«*;.-»5  .     .  (3) 

v  =  rco  Formula  XXVII.  .  (4) 

The  angular  velocity  of  a  point  is  its  linear  velocity  divided  by  the 
radius;  the  linear  velocity  is  the  angular  velocity  multiplied  by 
the  radius. 

Problems 

1.  The  circumference  of  a  flywheel  6  feet  in  diameter  is  moving  3000  feet 
per  minute.     What  is  the  angular  velocity  of  the  wheel? 

Ans.    to  =  16.67  radians    per  sec. 

2.  What  is  the  angular  velocity  of  a  pulley  which  is  making  300  revolutions 
per  minute?  Ans.    «  =  WTT  =  31.416  radians  per  sec. 

3.  A  steel  ball  is  whirled  in  a  vertical  plane  on  the  end  of  a  rod  4  feet  in 
length  at  the  rate  of  8  revolutions  in  4  seconds.     If  the  ball  is  released  from 
the  rod  when  the  rod  is  horizontal  and  is  moving  upward,  how  high  will  the 
ball  rise?     How  many  times  will  the  ball  rotate  on  its  axis  before  it  returns 
to  the  starting  point? 

The  formula  for  displacement  in  simple  harmonic  motion  is 

x  =  r  cos  JZg  t 
\  m 

The  expression  */  —  ^t  is  the  angular  displacement  in  the  circle 
of  reference,  and  .*/  —  -  is'  the  angular  velocity  in  that  circle. 

\     fit 

The  time  of  a  complete  period,  which  corresponds  with  the 
time  of  one  revolution  in  the  circle  of  reference,  is 

t.  =  2,  +  ,          -  2r  j. 


200.  Kinetic  Energy  of  Rotation.  —  Fig.  276  shows  an  element 
of  mass  dm  at  a  distance  r  from  an  axis  through  0.  If  this 
element  rotates  about  the  axis  with  an  angular  velocity  co,  its 
linear  velocity  is  rco,  and  its  kinetic  energy  is 

(3) 


336  MECHANICS  [ART.   200 

The  kinetic  energy  of  the  entire  mass  rotating  with  angular 
velocity  about  the  axis  through  0  is  the  integral  of  Equation  (1). 


dm;  (2) 

U  =  kinetic  energy  .=  •—-•        Formula  XXVIII 

If  the  mass  is  expressed  in  pounds  and  if 
g  =  32.174,  Formula  XXVIII  gives  the 
kinetic  energy  in  foot  pounds.  If  the  mass 
is  given  in  kilograms  and  g  =  9.81,  the  for- 
mula gives  the  kinetic  energy  in  kilogram- 
FIG.  276.  meters. 

Problems 

1.  A  homogeneous  solid  cylinder  4  feet  in  diameter  and  weighing  600 
pounds  revolves  on  its  axis  at  the  rate  of  2  revolutions  per  second.     Find 
its  kinetic  energy  in  foot  pounds. 

TT       600  X  2  X  16 X  9.8696      on._  ,,    „ 
AnS'   U  2  X  32.174 

2.  A  hollow  cylinder  6  feet  outside  diameter  and  4  feet  inside  diameter 
weighs  800  pounds  and  revolves  on  its  axis  at  the  rate  of  5  revolutions  per 
second.     Find   its   kinetic    energy.  Ans.     U  =  79,756    ft.  Ib. 

3.  A  flywheel  has  a  radius  of  gyration  of  1.8  feet  and  weighs  600  pounds. 
Find  its  kinetic  energy  when  it  is  making  300  revolutions  per  minute. 

Ans.     U  =   29,818  ft.  Ib. 

4.  The  flywheel  of  Problem  3  is  brought  to  rest  by  means  of  a  brake  which 
rubs  on  the  outer  surface.     The  diameter  of  the  wheel  is  4  feet.     The  brake 
exerts  a  normal  pressure  of  100  pounds  and  the  coefficient  of  friction  is 
0.24.     How  many  revolutions  will  the  wheel  make  before  it  comes  to  rest? 

Ans.     96.3  revolutions. 

6.  A  homogeneous  solid  cylinder  which  is  4  feet  in  diameter  and  weighs 
400  pounds  rotates  on  a  frictionless  axis.     A  rope  is  wound  several  times 
around  the  cylinder  and  is  fastened  to  it.     A  constant  pull  of  120  pounds 
is  applied  to  the  rope  while  the  cylinder  makes  5  revolutions.     What  is  the 
work  done  by  the  rope  on  the  cylinder?     What  is  the  final  velocity  of  the 
cylinder  at  the  end  of  5  revolutions? 
Ans.     U  =  120s   =  120  X  47r  X  5  =  7540  ft.  Ib.; 
^^  =  7540,  co  =  24.64  radians  per  sec. 

6.  In  Problem  5,  what  is  the  final  velocity  of  the  surface  of  the  cylinder 
in  feet  per  second.  v  =  49.28  ft.  per  sec. 


CHAP.  XXI]         ANGULAR  DISPLACEMENT 


337 


7.  A  mass  of  120  pounds  is  hung  on  the  rope  of  Problem  5,  Fig.  277. 
Find  the  angular  velocity  of  the  cylinder  and  the  linear  velocity  of  the 
rope  after  the  mass  has  moved  60  feet. 


20 
1280co2 


=  7200; 


w   =  19.0  radians  per  sec;  v  =  38.0  ft.  per  sec. 

8.  A  homogeneous  solid  cylinder  weighing  200  pounds  is  2  feet  in  diameter. 
This  cylinder  is  coaxial  with  a  second  cylinder  which  is  6  inches  in  diameter 
and  weighs  40  pounds.  A  rope  wound  on  the  smaller  cylinder  supports  a 
mass  of  60  pounds.  If  there  is  no  friction  and  if  the  mass  starts  from  rest, 
what  velocity  will  it  acquire  while  it  moves  100  feet? 

Ans.     v  =  15.16   ft.    per   sec. 


FIG.  277. 


FIG.  27S. 


9.  A  uniform  rod  AB  is  6  feet  in  length  and  is  hinged  at  A.     The  rod 
is  placed  in  the  position  of  unstable  equilibrium  and  then  released.     Find 
its  velocity  when  it  reaches  the  horizontal  position  of  Fig.  278. 

Ans.     Velocity  of  B  is  24.06  ft.  per  sec. 

10.  Solve  Problem  9  for  the  velocity  when  rod  is  30  degrees  below  the 
horizontal  and  when  the  rod  is  vertical  downward. 

Ans.     4.91  radians  per  sec.;  5.67  radians  per  sec. 

201.  Rotation  and  Translation. — In  the  preceding  article,  the 
axis  of  rotation  is  stationary  and  the  kinetic  energy  is  the  energy 
of  rotation.  When  a  body  is  rotating  about  an  axis  and  the  axis 
has  a  linear  motion,  the  total  kinetic  energy  is  the  sum  of  the 
energy  of  rotation  and  the  energy  of  translation. 


22 


?-£!  +  £. 

&Q  &Q 


(1) 


.338  MECHANICS  [ART.  202 

Since   energy   is   not   a   vector   quantity,  the  total  energy  is 
arithmetric  sum  of  the  two  terms. 

Figure  279  shows  a  wheel  of  radius  r  which  rolls  without 
slipping  on  a  horizontal  surface.  If  the  axis  were  stationary, 
the  velocity  of  the  circumference  of  the  wheel  relative  to  the 
point  B  would  be  rw.  When  the  wheel  does 
not  slip,  the  velocity  of  its  axis  relative  to  the 
surface  is  v  =  r<o. 
•  V  Problems 


1.  A    homogeneous    solid    cylinder    is    4    feet   in 
diameter  and  weighs  240  pounds.     It  is  rolling  at  a 
speed  of  60  feet  per  second.     Find  the  kinetic  energy 
279  °^  *rans^ati°n>  tne  kinetic  energy  of  rotation,  and  the 

total  kinetic  energy. 

Ans.     13,426  ft.-lb.;  6713  ft.-lb.;  20,739  ft.-lb. 

2.  When  a  homogeneous  cylinder  is  rolling  without  slipping  on  a  surface, 
show  that  the  kinetic  energy  of  rotation  is  one-half  the  kinetic  energy  of 
translation. 

3.  A  homogeneous  solid  cylinder  rolls  down  an  inclined  plane  which 
makes  an  angle  of  20  degrees  with  the  horizontal.     What  velocity  will  it 
acquire  while  moving  100  feet  from  rest?     What  velocity  would  it  acquire 
if  it  were  to  slide  down  the  plane  with  no  friction? 

Ans.    38.3  ft.  per  sec.;  46.9  ft.  per  sec. 

4.  In  Problem  3,  what  is  the  average  velocity  when  the  cylinder  rolls 
down  the  plane?     What  is  the  time  of  descent?        Ans.     t  =  5.22  sec. 

5.  The  outside  diameter  of  a  hollow  cylinder  is  4  feet  and  the  inside  diameter 
is  3.9  feet.     The  cylinder  rolls  100  feet  down  a  plane  which  makes  an  angle 
of  10  degrees  with  the  horizontal.     Find  the  final  velocity  of  the  hollow 
cylinder.     What  velocity  would  a  solid  cylinder  acquire  in  rolling  the  same 
distance  down  this  plane? 

Ans.  Velocity  of  hollow  cylinder  =  23.78  ft.  per  sec.;  velocity  of  solid 
cylinder  =  27.29  ft.  per  sec. 

6.  A  sphere  which  is  8  feet  in  diameter  rolls  on  two  parallel  rods  which 
are  1  foot  in  diameter.     The  axes  of  the  rods  are  5  feet  apart.     Find  the 
ratio  of  the  angular  velocity  of  the  sphere  to  the  linear  of  velocity  of  its 
center. 

7.  A  homogeneous  solid  sphere  rolls  on  a  surface.     What  is  the  ratio  of 
its  energy  of  rotation  to  its  energy  of  translation? 

202.  Translation  and  Rotation  Reduced  to  Rotation. — Trans- 
lation and  rotation  in  the  same  plane  may  be  regarded  as  rotation 
about  some  axis.  Figure  280  shows  a  cylinder  of  radius  r,  which 
is  rolling  with  velocity  v  on  a  plane  surface.  The  angular  velocity 

of  the  cylinder  about  its  axis  is  co  =  —     The  velocity  at  any 
point  may  be  found  from  the  resultant  of  the  velocity  of  the  axis 


CHAP.  XXI]         ANGULAR  DISPLACEMENT  339 

of  the  cylinder  and  the  angular  rotation  about  the  axis.  The 
velocity  at  any  point  may  also  be  found  from  the  angular  motion 
about  the  line  of  contact  with  the  plane.  In  Fig.  280,  the 
cylinder  may  be  regarded  as  rotating  about  an  axis  through  O 
perpendicular  to  the  plane  of  the  paper.  The  axis  of  the  cylinder 
moves  a  distance  vdt  during  a  time  in- 
terval dt.  The  angular  velocity  with 
respect  to  the  instantaneous  axis  through 

B  is  w  =  —     Since    the    cylinder    is    a 

rigid  body,  the  angular  velocity  of  all 
parts  is  the  same.  All  parts  of  the  cylin- 
der may,  therefore,  be  regarded  as  rotat- 
ing with  angular  velocity  about  the 
instantaneous  axis  through  B. 

The  kinetic  energy  of  a  body  which  rolls  on  a  plane  surface  is 


If  r  is  the  radius  from  the  axis  of  the  body  to  the  line  of  contact 
with  the  surface,  v  =  ro;  and  Equation  (1)  may  be  written 


(3) 


in  which  &o  is  the  radius  of  gyration  with  respect  to  the  axis 
through  the  center  of  gravity,  and  k  is  the  radius  of  gyration 
with  respect  to  the  line  of  contact  with  the  surface.  These 
equations  show  that  the  energy  computed  from  the  linear 
velocity  of  the  center  of  mass  and  the  angular  velocity  about 
that  center  is  the  same  as  the  energy  computed  from  the  angular 
velocity  about  the  line  of  contact. 

Problems 

1.  A  homogeneous  solid  cylinder  which  is  4  feet  in  diameter  and  weighs 
600  pounds  rolls  on  a  plane  surface  with  a  velocity  of  20  feet  per  second. 
Find  the  kinetic  energy  in  foot-pounds,  regarding  the  axis  as  moving  20 
feet  per  second,  and  the  cylinder  as  rotating  on  its  axis  at  the  rate  of  10 
radians  per  second.     Also  find  the  kinetic  energy,  regarding  the  cylinder 
as  rotating  about  the  axis  of  contact  with  the  surface  with  an  angular  velocity 
of  10  radians  per  second.  Ans.     U  =  5594.4  ft.-lb. 

2.  A  bar  of  length  I  is  rotating  with  angular  velocity  w  about  an  axis 
through  one  end  perpendicular  to  its  length.     Find  its  kinetic  energy, 


340 


MECHANICS 


[ART.   203 


3.  Solve  Problem  (2)  from  the  condition  that  the  center  of  gravity  of  the 
bar  has  a  linear  velocity  of  -£  and  the  bar  rotates  about  the  center  with  an 

angular  velocity  of  <o  radians  per  sec.  TT  _  mZ2o?2 

Ans.   u  —  — 7; — • 
60 

4.  In  Fig.  281,  find  the  direction  and  magnitude  of  the  velocity  of  each 
of  the  points  A,  B,  C,  D,  and  E.     Solve  by  regarding  the  axis  as  moving 
in  a  straight  line  while  the  points  revolve  about  it. 

5.  Solve  Problem  5  by  considering  the  cylinder  as  rotating  about  the  line 
of  contact  with  the  plane. 

6.  A  car  weighing  200  pounds  has  a  radius  of  gyration  of  2  feet  with 
respect  to  the  vertical  a:is  through  its  center  of  gravity.     The  car  moves 
with  a  velocity  of  40  feet  per  second  on  a  straight  track.     If  no  energy  is 
lost  in  friction,  what  is  its  velocity  when  it  runs  on  to  a  curve  of  10-foot 
radius?  Ans.     39.2  ft.   per  sec. 

A 


f  /        \0    \ © 


s.     A 


®\ 


FIG.  281. 


c£J@ 

B 
FIG.  282. 


Rotation  of  a  body  about  any  axis  with  an  angular  velocity 
of  co  radians  per  second  is  equivalent  to  a  linear  velocity  of 
rco  feet  per  second  combined  with  an  angular  velocity  of  co  radians 
per  second  about  a  parallel  axis  through  the  center  of  gravity. 
Figure  282  shows  four  positions  of  a  body  AB  which  rotates 
about  an  axis  0  normal  to  the  plane  of  the  paper.  The  axis 
through  the  center  of  gravity  of  AB  passes  through  C.  At 
position  1  at  the  right  of  the  figure,  the  point  B  is  to  the  right  of 
C.  At  the  second  position,  B  is  above  C.  At  the  third  position, 
B  is  to  the  left  of  C.  At  the  last  position,  B  is  below  C.  It  is 
evident,  therefore,  that  when  the  body  makes  one  revolution 
about  0,  it  makes  one  revolution  about  C. 

203.  Summary. — Angular  displacement  is  measured  in  radians. 
Angular  velocity  is  measured  in  radians  per  second.  When  the 
velocity  is  constant,  the  displacement  is  the  product  of  the  velocity 
multiplied  by  the  time.  When  the  velocity  varies,  the  displace 
ment  is  the  product  of  the  average  velocity  multiplied  by  the  time. 


CHAP.  XXI]         ANGULAR  DISPLACEMENT  341 

If  r  is  the  distance  from  a  point  to  the  axis  of  rotation,  the 
linear  velocity  of  the  point  due  to  an  angular  velocity  of  co  radians 
about  the  axis  is  given  by 

v  =  rco.  Formula   XXVII 

This  linear  velocity  is  normal  to  the  direction  of  r. 

If  the  axis  of  rotation  is  moving  with  a  linear  velocity,  the 
resultant  linear  velocity  of  any  point  is  obtained  by  combining 
the  linear  velocity  of  rotation  with  the  linear  velocity  of  the 
axis.  These  velocities  are  added  as  vectors. 

The  kinetic  energy  of  a  rotating  body  is 

U  =  ^  Formula  XXVIII 

20 

This  expression  for  kinetic  energy  corresponds  with  the  expression 
for  the  kinetic  energy  of  a  body  which  has  a  linear  velocity. 
Angular  velocity  replaces  linear  velocity,  and  moment  of  inertia 
replaces  mass  or  inertia. 

If  the  axis  of  rotation  is  moving  with  a  linear  velocity,  the 
total  energy  is  the  sum  of  the  energy  of  rotation  and  the  energy 
of  translation.  Energy  is  a  scalar  quantity;  kinetic  energy  is 
always  positive. 


CHAPTER  XXII 
ACCELERATION  TOWARD  THE  CENTER 

204.  Components  of  Acceleration. — In  Fig.  283,  a  point  at 
A  i  in  the  circumference  of  a  circle  of  radius  r  is  moving  with  a 
linear  velocity  v\.  When  the  point  reaches  the  position  A2,  its 
velocity  is  v^.  The  direction  and  magnitude  of  the  velocity  at  A  i 
is  given  by  the  tangent  AI  C;  the  direction  and  magnitude  of  the 
velocity  at  Az  is  given  by  the  tangent  A2D.  Figure  283,  II, 
is  the  vector  diagram  of  the  velocities  with  a  common  initial 
point  A.  The  vector  CD  of  Fig.  283,  II,  represents  the  change  of 
velocity  while  the  body  moves  from  A\  to  A2.  This  increment 


Vector 


Diagram 


II 


FIG.  283. 

of  velocity  may  be  resolved  into  two  components  CE  and  ED. 
The  length  AE  is  made  equal  to  AC.  The  component  ED  repre- 
sents the  increment  which  is  due  to  change  in  the  magnitude  of 
the  velocity;  the  component  CE  represents  the  increment  which  is 
due  to  change  in  the  direction.  Since  AEC  is  an  isosceles  triangle, 
the  direction  of  CE  is  normal  to  the  line  which  bisects  the  angle 
CAE,  and  is-  parallel  to  the  line  which  bisects  the  line  AiOAz  of 
Fig.  283,  I 

205.  Acceleration  Toward  the  Center. — If  v  is  the  magnitude 
of  the  velocity  at  A\  and  if  the  distance  from  A \  to  A 2  is  infinitesi- 
mal, the  increment  CE  is 

dv  =  vdO,  (1) 

342 


CHAP.  XXII]    ACCELERATION  TOWARD  CENTER         343 

in  which  d6  is  the  change  in  the  direction  of  the  velocity  and  dv 
is  the  increment  which  is  due  to  the  change  in  direction.  Since 
the  direction  of  the  increment  is  perpendicular  to  the  line  which 

dB 
makes  an  angle  —  with  the  direction  of  vi,  the  change  of  velocity 

m 

is  toward  the  center  of  the  circle  of  Fig.  283,  I. 
The  linear  acceleration  toward  the  center  is 


-  dv        v 
' 

Since  AiA2  =  r  dO  =  vdt, 


-        - 
'  Jt  ~     dt 


d0  _  v 
di  ~'~  r' 


Formula  XX 

Since  v  =  rco; 

a  =  rco2.  (3) 

Formula  XX  has  already  been  derived  in  Art.  162  for  motion 
in  a  circular  path  with  uniform  speed. 

Problems 

1.  A  body  is  whirled  in  a  circular  path  6  feet  in  diameter  with  a  speed  of 
15  feet  per  second.     Find  its  acceleration  toward  the  center  due  to  change  of 
direction.  Ans.     a  =  75  ft.  per  sec.  per  sec. 

2.  A  wheel  4  feet  in  diameter  is  making  5  revolutions  per  second.     What  is 
the  acceleration  toward  the  center  of  a  point  on  the  rim  of  the  wheel? 

Ans.     a  =  2  X  102  X  7T2  =  1974  ft.  per  sec.  per  sec. 

3.  A  wire  is  bent  into  the  form  of  an  ellipse  the  equation  of  which  is 


A  body  moves  on  this  wire  with  a  constant  speed  v.     Find  the  acceleration 

of  the  body  when  it  is  at  the  end  of  the  major  axis.  vza 

J  Ans.     a  =  T-Z-. 

62 

Problem  3  may  be  solved  by  substituting  the  radius  of  curvature  of  the 
ellipse  in  Formula  XX.  Another  solution,  which  does  not  depend  upon 
Formula  XX,  is  given  below.  Differentiating  the  equation  of  the  ellipse 
with  respect  to  t, 

x_dx  +  y_dy  =  Q  (4) 

a2  dt  +  62  dt 

When  x  =  a,   y  =  0,    Substitution   in   E  uation    (4)    shows  that  ^r  =  0 
when  x  =  a.     Since  the  X  component  of  the  velocity  is  zero  at  this  position, 


344 


MECHANICS 


[ART.   206 


the  Y  component  must  equal  v;  -£  =  v  when  x  =  a.     The  motion  at  the 

point  A  of  Fig.  284  is  vertically  upward. 

Differentiating  Equation  (4)  with  respect  to  t, 

x_  d^x      _1  /dx\  2      y_  d^y      1^  fdy\  2  =  0  ,-v 

a2  d<2       a2  Vft  /         62  <ft2       fe2-  \d«  / 

From  the  vector  triangle  of  velocities, 
V 


Differentiating  Equation  (6)  with  v  constant, 

l^+t^-ft          (7> 


When  x  =  a,-rr 


FIG.  284. 


0.     Substitution  in  Equa- 

d2!/ 

tion  (7)  show  that  -^   is   zero:    hence   the 
2 


acceleration  has  no  vertical  component  when 

x  =  a.  Substituting  the  conditions  that  the  horizontal  component  of  the 
velocity  and  the  vertical  component  of  the  acceleration  are  each  zero  when 
x  =  a,  Equation  (5)  becomes 


_ 

°' 

d*x      a  (dy\  *      av* 

dt*       b*  \dt  /         Z>2 

If  a  =  b,  the  ellipse  is  a  circle  and  Equation  (9)  reduces  to  Formula  XX. 

206.  Centrifugal  Force.  —  When  a  particle  is  rotating  about 
an  axis,  the  acceleration  toward  the  axis  which  results  from  the 
change  of  direction  is 


Since  the  force  which  produces 
the     force 


,       ,.        .ma 
acceleration   is  — , 
y 


which    produces    this    accelera- 
tion in  a  particle  of  mass  dm  is 

dmv*       dmrw2 

-  or This   is   called 


the  centrifugal  force.     The  force  0  ^ x ^ 

on   the   particle  is  toward   the  Fio  2g5 

axis;  it  is,  therefore,  centripetal 

force.     The  force   which   the   rotating   particle   exerts   on   the 

body  or  surface  which  holds  it  to  a   curved  path  is  directed 

away  from  the  axis;  it  is,  therefore,  centrifugal. 

All  parts  of  a  small  particle  are  at  the  same  distance  from 


CHAP.  XXII]    ACCELERATION  TOWARD  CENTER         345 

the  axis.  When  a  body*  of  considerable  dimensions  is  rotating 
about  an  axis,  it  is  necessary  to  find  an  expression  for  the  resultant 
centrifugal  force  exerted  by  the  entire  body.  Figure  285  repre- 
sents a  body  which  is  rotating  about  an  axis  at  0  normal  to  the 
plane  of  the  paper.  The  element  of  mass  dm  at  a  distance 
r  from  the  axis  is  subjected  to  a  pull  toward  the  axis.  The 
magnitude  of  this  pull  is 

dp  =  r-^  CD 

To  find  the  resultant  pull  on  all  the  elements  of  mass,  this  force 
is  resolved  into  components  parallel  to  the  coordinate  axes. 
The  resolution  parallel  to  the  X  axis  is 

dPx  =  —  r  cos  6  dm  =  —  xdm\  (2) 

n        co2  r    ,          w*mx  /ON 

Px  =  —  I  xdm  = (3) 

9  J  Q 

In  a  similar  way,  the  component  parallel  to  the  Y  axis  is  found  to 

P.-*&.  •  (4) 

I/ 

The  resultant  force  is 

~         wfco2        mV*  _,  , 

P  =  =  — i  Formula  XXIX 

Q  ffr 

in  which  f  is  the  distance  of  the  center  of  gravity  of  the  mass 
from  the  axis,  and  v  is  the  linear  velocity  of  the  center  of  gravity. 

Problems 

1.  A  mass  of  4  pounds  is  whirled  in  a  horizontal  plane  at  the  end  of  a  rod 

4  feet  in  length.     The  velocity  of  the  center  of  gravity  of  the  body  is  20 
feet  per  second.     What  tension  does  the  body  exert  on  the  rod? 

Ans.     P  =  12.43  Ib. 

2.  The  body  of  Problem  1  makes  2  revolutions  per  second  about  a  vertical 
axis.     Find  the  tension  in  the  rod.     Use  the  formula  in  terms  of  the  angular 
velocity.  ,  Ans.     P  =  78.53  Ib. 

3.  A  mass  of  5  pounds  is  whirled  in  a  vertical  plane  at  the  end  of  a  cord 

5  feet  in  length.     Find  the  tension  at  the  top  and  bottom  of  its  path  if  the 
velocity  at  each  position  is  24  feet  per  second. 

Ans.     12.90  Ib.  at  the  top;  22.90  Ib.  at  the  bottom. 

4.  In  Problem  3,  the  velocity  at  the  top  is  24  feet  per  second.     If  no  force 
acts  except  gravity,  find  the  velocity  at  the  bottom  and  the  tension  in  the 
cord. 

5.  A  pail  of  water  is  whirled  in  a  vertical  plane  about  a  horizontal  axis. 


346  MECHANICS  [ART.  207 

The  distance  from  the  axis  to  the  surface  of  the  water  is  3  feet.     Find  the 
minimum  velocity. 

6.  In  a  "Loop  the  Loop,"  the  radius  of  curvature  at  the  top  is  20  feet. 
How  much  must  the  starting  point  be  elevated  above  the  top  of  the  loop 
if  no  allowance  is  made  for  loss  of  energy  which  is  due  to  friction,  or  for  the 
kinetic  energy  of  rotation? 

7.  In  a  "Loop  the  Loop,"  the  starting  point  is  40  feet  above  the  lowest 
point.     The  radius  of  curvature  at  the  lowest  point  is  40  feet.     What  will 
be  the  weight  on  a  spring  balance  of  a  200-pound  body  when  passing  the 
lowest  point?  Ans.     P  =  600  Ib. 

8.  A  boy  on  a  bicycle  makes  a  turn  of  40  feet  radius  at  a  speed  of  20  feet 
per  second.     How  much  must  he  lean  from  the  vertical?     Ans.     17°  16'. 

9.  If  the  boy  of  Problem  8  makes  the  turn  on  a  level  street,  what  must  be 
be  the  minimum  value  of  the  coefficient  of  friction?     Ans.    f  =  0.311. 

10.  If  the  distance  between  rails  of  a  track  is  56.5  inches,  how  much 
must  the  outer  rail  be  elevated  for  a  curve  of  800  feet  radius  in  order  that 
the  resultant  force  may  be  normal  at  a  speed  of  30  miles  per  hour? 

Ans.     4.25  inches. 

11.  A  dynamo  armature  weighing  400  pounds  runs  at  1200  revolutions 
per  minute.     If  its  center  of  gravity  is  0.03  inch  from  its  axis,  what  is  the 
maximum  vertical  pull  upward  on  the  bearings,  and  what  is  the  maximum 
horizontal  force  on  the  bearings?  Ans.     90.8  Ib.  490.8  Ib. 

207.  Static  Balance. — The  armature  of  Problem  11  of  the 
preceding  article  is  "out  of  balance."  The  resultant  pressure 
downwards  when  the  center  of  gravity  is  below  the  axis  is  890 
pounds.  The  resultant  pressure  upward,  when  the  center  of 
gravity  is  above  the  axis,  is  90  pounds.  The  increased  pressure 
which  is  due  to  centrifugal  force  bends  the  shaft.  This  deflection 
of  the  shaft  increases  the  distance  of  the  center  of  gravity  from 
the  axis  of  rotation  and  still  further  increases  the  centrifugal 
force.  If  the  time  of  vibration  of  the  shaft  should  coincide  with 
the  time  of  one  revolution,  the  deflection  and  stress  would  be 
increased  still  more.  Even  if  the  condition  of  resonance  does  not 
exist,  the  additional  stresses  which  are  due  to  the  centrifugal 
force  are  dangerous  and  destructive.  It  is  evident,  therefore 
that  rapidy  running  machinery  must  be  accurately  balanced  so 
that  the  center  of  gravity  of  the  rotating  part  may  coincide  with 
its  axis  of  rotation. 

In  Problem  11,  the  moment  of  the  mass  about  the  axis  is 
400  X  0.03  =  12  inch  pounds.  This  moment  may  be  balanced 
by  a  mass  of  6  pounds  at  2  inches  from  the  axis  or  a  mass 
4  pounds  at  3  inches  from  the  axis. 

The  static  balance  of  an  armature  or  other  body  designed 
to  rotate  at  high  speed  is  frequently  determined  by  placing  the 


CHAP.  XXII]    ACCELERATION  TOWARD  CENTER         347 

axle  on  a  pair  of  horizontal  rails.  The  axle  rolls  on  these  rails 
until  the  center  of  gravity  is  vertically  below  the  axis.  Addi- 
tional load  is  then  placed  directly  above  the  axle  to  secure  a 
balance.  When  the  axle  is  in  equilibrium  at  any  position,  the 
resultant  center  of  gravity  is  known  to  be  on  the  axis.  The 
body  is  then  in  static  balance. 

208.  Running  Balance. — A  shaft  may  be  in  perfect  static 
balance  and  yet  when  running,  be  decidedly  out  of  balance. 
For  perfect  running  balance,  the  shaft  and  the  bodies  which 
it  carries  must  be  symmetrical  with  respect  to  the  axis.  In  Fig. 


w/////$w//, 
R. 

i 

p1 

t 

I 

£©W 

II 

\      \ 

FIG.  286. 

* 

286,  I,  the  mass  W  is  at  a  distance  a  from  the  axis.     A  second 
mass  W  at  a  distance  b  from  the  axis  on  the  opposite  side  will 
produce  a  static  balance  if 

W'b  =Wa. 

In  Fig.  286,  II,  P'  represents  the  centrifugal  force  which  is  due 
to  W,  and  Pin  the  opposite  direction  represents  the  equal  centrifu- 
gal force  which  is  due  to  W.  Since  these  forces  are  not  along 
the  same  line,  they  produce  bending  moments  in  the  shaft  and 
varying  reactions  in  the  bearings. 

Example 

In  Problem  11  of  the  preceding  article,  the  axle  is  60  inches  long  from 
center  to  center  of  bearings.  The  center  of  mass  of  the  armature  is  32 
inches  from  the  left  bearing.  Static  balance  is  secured  by  the  addition  of 
2.4  pounds  of  lead  inside  a  collar  located  10  inches  from  the  right  bearing. 
The  center  of  gravity  of  the  lead  is  5  inches  from  the  axis.  Find  the  reactions 
of  the  bearings  when  the  armature  is  making  1200  revolutions  per  minute. 

The  centrifugal  force  of  the  mass  of  the  armature  is  490.8  pounds.  The 
centrifugal  force  of  the  2.4  pounds  of  lead  is  also  490.8  pounds.  In  Fig. 

287,  these  centrifugal  forces  are  horizontal.     The  horizontal  reactions  of 
the  bearings  are  HI  and  H2.     By  moments  about  the  vertical  lines  through 


348  MECHANICS  [ART.  209 

the  bearings,  HI  is  found  to  be  147.2  pounds  in  one  direction  and  H2  is 
found  to  be  an  equal  force  in  the  opposite  direction.  When  the  center  of 
gravity  is  below  the  axis,  the  centrifugal  force  adds  147.2  pounds  to  the 


60" "- 

1490.8* 


•-- 32-- > 


490.8' 

C. 

FIG.  287. 

load  on  the  left  bearing  and  subtracts  the  same  amount  from  the  load  on  the 
right  bearing. 

Problems 

1.  Find  the  reactions  of  the  bearings  of  the  above  example  when   the 
center  of  gravity  of  the  armature  is  directly  below  the  axis. 

Ans.     334.3  lb.;  65.8  Ib. 

2.  Solve  Problem  1  when  the  center  of  gravity  of  the  armature  is  directly 
above  the  axis  of  the  shaft. 

209.  Ball  Governor. — In  Fig.  288,  AB  is  shaft  which  rotates 
about  a  vertical  axis,  and  C  is  a  heavy  ball  connected  to  the 
shaft  by  a  rod.  The  rod  is  attached  to  the  shaft  at  B  by  means 
of  a  hinge  which  permits  the  rod  and  the  ball  C  to  rotate  in  a 
vertical  plane.  As  the  vertical  shaft  rotates,  the  centrifugal 
force  in  C  develops  a  moment  which 
tends  to  lift  BC  toward  a  horizontal 
position. 

If  6  is  the  angle  which  BC  makes 
with  the  vertical,  if  I  is  the  length  of  BC 
from  the  hinge  to  the  center  of  gravity, 
and  if  m  is  the  combined  mass  of  the  ball 
and  the  rod,  the  centrifugal  force  is 

u       ml  sin  0co2 

ti  =  — •  (1) 

FIG.  288.  Q 

The  moment  arm  of  this  horizontal  centrifugal  force  about  the 
hinge  at  B  is  I  cos  8  while  the  moment  arm  of  the  weight  is  I 
sin  6.  Equating  the  moment  of  the  centrifugal  force  to  the 
moment  of  the  weight, 

cos  e  =  ^  (3) 


•»* 

•  u*'  -    ^ , 

CHAP.  XXII]    ACCELERATION  TOWARD  CENTER         349 

Problems 

1.  The  length  of  a  governor  arm  is  12  inches.     At  what  speed  will  the 
arm  stand  at  60  degrees  with  the  vertical? 

Ans.     8.02  radians  per  sec.;  76.6  rev.  per  min. 

2.  In  Problem  1,  what  is  the  value  of  6  when  the  speed  is  50  revolutions 
per  minute?  Ans.     6  =  0. 


FIG.  289. 


FIG.  290. 


3.  Figure  289  shows  several  arms  of  different  lengths  which  are  attached 
to  the  same  hinge.     Show  that  the  center  of  gravity  of  each  will  fall  on  the 
same  horizontal  line,  provided  the  angular  velocity  is  sufficient  to  make  all 
of  them  leave  the  vertical  position. 

4.  In  Problem  1,  find  the  angle  for  angular  velocities  of  6,  7,  8,  9,  and  10 
radians  per  second.     Ans.     26°  40',  48°  58',  59°  49',  66°  36',  and  71°  14'. 

5.  In  Problem  1,  find  the  angular  velocity  for 
angles  of  30°,  40°,  50°,  60°,  and  70°. 

'  Ans.     6.10,  6.48,  7.07,  8.02,  and  9.70 
radians  per  sec. 

6.  In  Fig.  290,  find  the  angular  velocity  for 
angles  of  30°,  40°,  50°,  60°,  and  70°. 

Ans.     4.72,  5.26,  5.90,  6.82,  and  8.33 
radians  per  sec. 

7.  Figure  291  shows  a  crossed-arm  governor 
such  as  is  used  on  some  chronographs.     Find 
the  expression  for  the  angular  velocity  in  terms 
of  the  angle  6,  the  length  a,  and  the  length  I. 

,          g  tan  0 

Ans.     or  =  -j—^ — ^ 

I  sin  0  —  a 

If  I  in  Fig.  291  is  12  inches  and  a  is  4  inches, 
find  the  value  of  the  angular  velocity  for  angles 
of  45°,  50°,  60°,  and  70°. 

Ans.     9.28,    9.41,    10.23,    and 


FIG.  291. 


12.07    radians    per    sec. 


The  form  of  governor  shown  in  Fig.  291  is  extremely  sensitive 
when   6  is  about  45  degrees.     A  change  of  angular  velocity  of 


350 


MECHANICS 


[ART.   210 


less  than  2  per  cent  causes  a  change  in  the  angle  from  45°  to 
50°. 

210.  Loaded  Governor. — Figure  292  shows  a  loaded  governor. 
In  this  figure,  the  length  of  the  links  which  connect  the  load  with 
the  arms  is  equal  to  the  distance  from  the  top  of  the  arm  to 
the  connection  with  the  link.  The  link  and  the  arm,  therefore, 
make  equal  angles  with  the  vertical.  The  centrifugal  force  is 

ml  sin  0co2 


H 


(1) 


FIG.  292. 


If  Q  is  the  tension  in  one  link  and  W  is  the  weight  lifted  by  both 
links, 


W 
Q  cos  6  =  -- 


The  horizontal  component  of  Q  is 


W 


N  =  Q  sin  0  =        tan  6. 


(2) 
(3) 


Consider  the  right  arm  and  the  right  link  as  a  system  in  equi- 
librium, and  take  moments  about  the  top  of  the  arm.  The  force 
from  the  weight  to  the  link  may  be  resolved  into  a  vertical 
component  downward  and  a  horizontal  component  N  directed 
toward  the  left.  The  moment  arm  of  the  vertical  component  is 
zero.  The  moment  arm  of  the  horizontal  component  is  2a  cos  6. 


W 
ml  sin  0  +  —  2  a  tan  6  cos  6  = 

m£2co2 
ml  +  TFa  =  - 


ml  sin  0co2 

g 
cos  e. 


I  cos  6; 


(4) 
(5) 


CHAP.  XXII]    ACCELERATION  TOWARD  CENTER         351 


Problems 

1.  In  Fig.  292,  the  length  Z  is  20  inches  and  the  length  a  is  12  inches. 
The  load  W  is  12  pounds  and  the  weight  of  each  arm  is  10  pounds.     Find 
the  speed  when  the  arms  make  an  angle  of  45  degrees  with  the  vertical. 

Ans.    6.85  radians  per  sec. 

2.  What  is  the  angular  velocity  required  to  lift  the  weight  of  Problem  1 
a  distance  of  2  inches  from  the  position  at  which  the  arms  make  an  angle  of 
45   degrees  with  the  horizontal?  Ans.     7.30  radians  per  sec. 

211.  Flywheel  Stresses. — When  a  wheel  is  rotating,  every  part 
of  its  mass  exerts  a  pull  due  to  the  centrifugal  force.    In  calculat- 
ing the  stresses,  the  rim  is  often 
considered    separately    and   the 
mass  of  the  spokes  is  neglected. 
The  force   on   a  rim  is  similar 
to  the  force  exerted  on  a  cylin- 
drical drum  from  the  pressure  of 
confined  liquid  or  gas.     The  cen- 
trifugal force  on  an  element  of 

rco2 
mass  of  the  rim  is  — .    The  rim 

g 

of  Fig.  293  may  be  considered  as 

made  up  of  circular  hoops  each 

of  which  is   1  inch  wide  and  t 

inches  thick.     If  w  is  the  weight 

of  a  rod  of  the  material  1  inch 

square  and  1  foot  long,  the  cen-  FlG  293 

trif ugal  force  per  foot  of  the  hoop 

is .     To  find  the  total  force  on  one  half  of  the  hoop  which 

tends  to  rupture  it  at  the  sections  A  and  B,  the  forces  on  this 
half  are  resolved  normal  to  the  diameter  through  A  and  B. 

The  sum  of  all  the  components  is  equal  to  a  force  of per 

foot  of  length  of  the  diameter. 


AV  „ 

Total  pull  = 


n 
2r  = 


Q 


(1) 


This  pull  is  resisted  by  the  tensile  strength  of  the  material  at  A  and 
B.     The  total  cross  section  is  2t.     If  St  is  the  unit  tensile  stress, 


2tSt  = 


(2) 
(3) 


352  MECHANICS  [ART.  212 

The  weight  of  a  rod  of  steel  1  inch  square  and  1  foot  long  is 
3.4  pounds.  The  weight  of  a  rod  of  cast  iron  is  about  3.2  pounds. 

Problems 

1.  The  rim  of  a  steel  wheel  is  moving  with  a  speed  of  80  feet  per  second. 
Find  the  tension  due  to  the  centrifugal  force.    Ans.    St  =  677  Ib. /in.2 

2.  A  steel  wheel  6  feet  in  diameter  is  making  300  revolutions  per  minute. 
Find  the  tension  in  the  rim  due  to  rotation.     Ans.    St  =  939  lb./in.2 

3.  If  the  density  of  cast  iron  is  90  per  cent  the  density  of  steel,  and  if  the 
allowable  tensile  stress  is  2000  pounds  per  square  inch,  what  is  the  maximum 
allowable  rim  velocity  for  cast  iron  wheels? 

Equations  (2)  and  (3)  may  be  derived  by  calculating  the 
centrifugal  force  on  one  half  of  the  rim.  If  the  breadth  is  b  and 
the  thickness  is  t,  the  volume  of  one-half  the  rim  is  btirr.  If  r  is 
in  feet  and  b  and  t  are  expressed  in  inches,  the  mass  is  wbtirr.  The 

centrifugal  force  is  —         — ,  in  which  r  is  the  distance  of  the  center 

y 
of  gravity  of  the  semicircular  rim  from  the  axis.     If  the  thickness 

rs  small,  r  =  — .     Substituting  this  value  of  r  and  equating  to 
the  tensile  stress. 


=  2btSt', 

o 

g   =  wrV  =  wv2 
Q  9  ' 

If  w'  is  the  density  in  pounds  per  cubic  inch, 


212.    Summary 

A  rotating  body  may  have  two  accelerations.  One  of  these 
accelerations  is  due  to  the  change  of  direction.  This  accelera- 
tion is  directed  toward  the  axis  of  rotation.  Its  magnitude  is 


-    =    r,.,2 


The  centrifugal  force  exerted  by  a  particle  of  m  pounds  mass  is 

ru 
9 


_  mv2 


CHAP.  XXII]    ACCELERATION  TOWARD  CENTER          353 

The  centrifugal  force  exerted  by  a  body  of  some  magnitude  is 

mv2       rarco2 
P  =  -=r-  = •  Formula  XXIX 

r  9  9 

in  which  r  is  the  distance  of  the  center  of  gravity  of  the  mass  from 
the  axis  of  rotation  and  v  is  the  linear  velocity  of  the  center  of 
gravity. 

A  rotating  body  is  in  static  balance  when  the  center  of  gravity 
coincides  with  the  axis  of  revolution.  A  body  is  in  running 
balance  when  the  center  of  gravity  of  every  small  portion  between 
two  planes  normal  to  the  axis  falls  on  the  axis. 

In  a  Watt  or  ball  governor,  the  arms  take  positions  of  equilib- 
rium under  the  action  of  the  horizontal  centrifugal  force  which 
varies  as  the  square  of  the  angular  velocity  and  as  the  distance 
from  the  axis,  and  the  constant  vertical  weights  of  the  arms  and 
the  additional  loads. 

Allowance  must  be  made  for  stresses  which  are  due  to  cen- 
trifugal force  in  high  speed  machinery.  Since  centrifugal  force 
varies  as  the  square  of  the  angular  velocity,  a  relatively  small 
increase  of  speed  may  become  a  source  of  danger. 


23 


CHAPTER  XXIII 
ANGULAR  ACCELERATION 

213.  Angular  Acceleration.  —  The  rate  of  change  of  angular 
velocity  is  the  angular  acceleration. 


a  = 


in  which  a  is  the  angular  acceleration.  Angular  acceleration  is 
expressed  in  radians  per  second  per  second.  For  small  intervals 
of  time,  the  angular  acceleration  is 

c&o       d*6 


Problems 

1.  A  fly  wheel  is  making  2  revolutions  per  second,  after  an  interval  of 
4  seconds,  it  is  making  5  revolutions  per  second.     Find  the  angular  accelera- 
tion.    By  means  of  the  average  velocity,  find  the  number  of  revolutions 
during  the  4-second  interval. 

Ans.     a  =  4.712  radians  per  sec.  per  sec.;  14  revolutions. 

2.  A  point  on  the  rim  of  a  wheel  6  feet  in  diameter  is  moving  with  a 
velocity  of  60  feet  per  second.     After  2  seconds,  it  is  moving  with  a  velocity 
of  90  feet  per  second.     Find  the  angular  acceleration  of  the  wheel. 

Ans.     5  radians  per  sec.  per  sec. 

3.  A  pencil  held  for  O.-l  second  against  the  rim  of  a  wheel  which  is  8  feet 
in  diameter  makes  a  mark  18  inches  long.     Two  seconds  later  the  pencil  is 
held  against  the  wheel  for  the  same  time  and  makes  a  mark  30  inches  long. 
Find  the  angular  acceleration  of  the  wheel. 

Ans.     a  =  1.25  radians  per  sec.  per  sec. 

214.  Displacement  with  Constant  Acceleration.  —  The  expres- 
sions for  angular  velocity  and  displacement  when  the  acceleration 
is  constant  are  similar,  to  those  for  linear  displacement  and 
velocity  when  the  linear  acceleration  is  constant. 

a)  =  co0  +  at  (1) 

e  =  0,0*  +  ^  (2) 

Equation  (1)  may  be  derived  from  the  definition  of  acceleration. 

354 


CHAP.  XXIII]      ANGULAR  ACCELERATION  355 

Equation  (2)  may  be  derived  by  multiplying  the  average  an- 
gular velocity  by  the  time.  Both  equations  may  be  derived  by 
integration. 

Problems 

1.  A  wheel  has  an  angular  acceleration  of  2  revolutions  per  second  per 
second.     Its  initial  velocity  is  3  revolutions  per  second.     What  will  be  its 
velocity  after  4  seconds?  Ans.     11  revolutions  per  second. 

2.  How  many  revolutions  will  the  wheel  of  Problem  1  make  during  the 
interval  of  4  seconds.     Solve  by  means  of  the  average  velocity. 

Ans.     26  revolutions. 

3.  A  wheel  is  making  1600  revolutions  per  minute.     After  an  interval  of 
20  seconds,  it  is  making  600  revolutions  per  minute.     Find  its  acceleration 
in  radians  per  second  per  second. 

Ans.  a  =  —  5.24  radians  per  sec.  per  sec. 

4.  How  many  revolutions  does  the  wheel  of  Problem  3  make  during  the 
interval  of  20  seconds?  Ans.     367  rev. 

215.  Acceleration  and  Torque. — In  Fig.  294,  a  mass  dm 
rotates  about  an  axis  at  0.  The  angular  acceleration  is  a  and 
the  distance  of  the  element  from  the  axis  is  r.  The  linear  accelera- 
tion of  the  element  of  mass  is  a  =  ra.  The 
force  which  must  be  applied  to  produce  this 
linear  acceleration  is 

dP  =        m-  (1) 

Q 

The  moment  of  this  force  with  respect  to  the 
axis  through  0  is 

,m       r2adm       a  »,  /ox 

dT  =  -       -  =  -r*dm.  (2)  FIG.  294. 

g  g 

The  moment  required  to  produce  an  angular  acceleration  in 
a  mass  made  up  of  elements  dm  is 

T  =  -  I  r*dm  =  — •  Formula  XXX 


g 

The  moment  with  respect  to  an  axis  of  a  rotating  body  is  called 
torque.  Torque  is  represented  algebraically  by  the  letter  T. 

Formula  XXX  for  angular  acceleration  is  similar  to  Formula 
XXII  for  linear  acceleration.  Force  in  Formula  XXII  is 
replaced  by  moment  of  force,  and  mass  (or  inertia)  is  replaced  by 
moment  of  inertia. 

If  force  is  expressed  in  absolute  units,  Formula  XXX  becomes 

T  =  la  (3) 

Equation  (3)  is  sometimes  given  as  the  physical  definition  of 


356 


MECHANICS 


[ART.    215 


moment  of  inertia.  The  moment  of  inertia  of  a  body  is  numeri- 
cally equal  to  the  moment  of  the  force  (in  absolute  units)  which 
gives  to  the  body  an  acceleration  of  1  radian  per  second  per 
second. 

Problems 

1.  A  homogeneous  solid  cy Under  is  4  feet  in  diameter  and  weighs  240 
pounds.  A  rope  to  which  a  force  of  60  pounds  is  applied  is  wound  several 
times  around  the  cylinder.  What  is  the  torque? 
What  is  the  angular  acceleration?  What  is  the  angu- 
lar velocity  after  4  seconds  and  what  angle  is  turned 
through  in  4  seconds  if  the  cylinder  starts  from 
rest? 

Ans.  T  =  120  ft.-lb.;  a  =  8.044  radians  per  sec. 
per  sec.  :o>  =  32. 176  radiance  per  sec.  ;0  =  64.35  radians. 
2.  Figure  295  shows  the  cylinder  of  Problem  1  with 
a  mass  of  60  pounds  hung  on  the  free  end  of  the 
rope.  Find  the  acceleration,  the  tension  in  the  rope, 
and  the  angular  velocity  after  4  seconds. 

If  P  is  the  tension  in  the  rope,  the  effective  force  on 
the  mass  of  60  pounds  is  60  —  P.  When  the  60-pound 
mass  is  considered  as  a  free  body  with  a  linear  acceleration  a,  Formula 
XXII  gives 

60-P  =  ™a.  (4) 


340  I  b, 


60  Ib. 

I 

FIG.  295. 


The  torque  on  the  cylinder  is  2P  and  Formula  XXX  gives 

2p  =    480« 


(5) 


.A  B 


Substituting  a  =  2a  in  Equation  (4)  and  eliminating  P  between  Equations 
(4)  and  (5) 

a  =  |  =  5.362  radians  per  sec.  per  sec.; 

P  =  40  Ib. 

One-third  of  the  weight  of  the  60- 
pound  mass  is  used  to  accelerate  the 
mass  itself  and  the  remaining  40  pounds 
accelerates  the  cylinder. 

3.  In   Problem   2  find  the   distance 
traversed  by  the  mass  of  60  pounds 
during  4  seconds.     Find  the  work  done 
by  gravity  and  equate  with  the  total 
kinetic  energy. 

Ans.  Total  work  =  5147.8  ft.-lb. 

4.  A   homogeneous  solid  cylinder  rolls  without  slipping  down  20-degree 
inclined  plane.     Find  the  acceleration.     Solve  by  considering  the  cylinder 
as  rotating  about  its  line  of  contact  with  the  plane. 

20  sin  20°      7.39     ,. 
Ans.     a  = ^ = radians  per  sec.  per  sec.; 

tjT  T 

a    =  7.34  ft.  per  sec.  per  sec. 


30 


FIG.  296. 


CHAP.  XXIII]       ANGULAR  ACCELERATION 


357 


6.  A  uniform  bar  AB,  Fig.  296,  6  feet  long  and  weighing  12  pounds,  is 
hinged  at  A.  What  is  the  angular  acceleration  when  the  bar  is  in  the  posi- 
tion at  which  it  makes  an  angle  of  30  degrees  with  the  horizontal? 

Ans.  a.  =  3.483  radians  per  sec.  per  sec. 
6.  Solve  Problem  5  for  the  horizontal  position  of  the  bar. 

Ans.     a  =  8.044  radians  per  sec.  per  sec. 

216.  Equivalent  Mass. — Problem  2  of  the  preceding  article 
may  be  solved  by  first  finding  the  mass  which  would  have  the 
same  velocity  as  the  rope  and  the  same  moment  of  inertia  as 
the  cylinder.  If  all  the  mass  of  the  'cylinder  were  concentrated 
at  the  surface,  it  would  move  with  the  linear  velocity  of  the  rope. 
The  moment  of  inertia  of  a  solid  cylinder  is 

/  =  ^-, 


W/"         ///  •// • ,.  ; 
W|  120  ib/  {/I 


II 


60  Ib, 


FIG.  297. 


m 


a  mass  -^-,  concentrated  at  the  surface  in  the  form  of  a  very 
A 

thin  hollow  cylinder  of  radius  r,  has  the  same  moment  of  inertia 
as  the  actual  solid  cylinder  of  mass  m.  The  problem  of  accelerat- 
ing a  solid  cylinder  of  mass  m  by  means  of  a  rope  to  which  a  mass 
m'  is  attached  is  equivalent  to  that  of  accelerating  a  thin  hollow 

cylinder  of  mass  -^-     The  problem  of  accelerating  a  thin  hollow 
& 

cylinder  is  equivalent  to  the  problem  of  Fig.  297,  II.  The 
mass  of  240  pounds  of  Problem  2  of  Art.  215  may  be  replaced 
by  a  mass  of  120  pounds  on  a  horizontal  plane  as  in  Fig.  297,  II 
and  the  problem  may  be  solved  as  a  problem  of  linear  accelera- 
tion. Since  the  effective  force  on  the  system  is  60  pounds  and 
the  total  equivalent  mass  is  180  pounds,  it  is  evident  that  the 
linear  acceleration  is  gr/3  and  that  one-third  of  the  force  of  60 
pounds  is  required  to  accelerate  its  own  mass. 


358  MECHANICS  ~,  [ART.  217 

Problems 

1.  A  wheel  is  6  feet  in  diameter  and  weighs  400  pounds.     Its  radius  of 
gyration  is  2.4  feet.     What  mass  concentrated  at  the  rim  will  have  the  same 
moment  of  inertia?  Ans.  256  1-b. 

2.  A  wheel  is  5  feet  hi  diameter  and  weighs  600  pounds.     Its  radius  of 
gyration  is  2  feet.     A  J^-inch  rope  is  wound  around  the  wheel  and  a  mass  of 
80  pounds  is  applied  to  the  free  end.     Find  the  acceleration  and  the  tension 
on  the  rope. 

3.  The  wheel  of  Problem  2  turns  on  a  4-inch  axle.     The  coefficient  of 
friction  is  0.04.     Find  the  acceleration  and  the  tension  on  the  rope. 

4.  A  cylinder  4  feet  in  diameter  and  weighing  200  pounds  is  coaxial  with 
an  axle  of  the  same  material  which  is  6  inches  in  diameter.     The  part  of  the 
axle  which  projects  from  the  cylinder  weighs  100  pounds.     A  1-inch  rope  is 
wound  round  the  axle  and  a  mass  of  200  pounds  is  hung  on  it.     If  there  is 
no  friction,  what  is  the  acceleration  of  the  rope  and  what  is  the  tension  in  it? 

217.  Reaction  of  Supports.  —  Figure  298  represents  a  body  which 

is  free  to  rotate  in  the  plane  of  the  paper  about  a  hinge  at  0. 

The  center  of  mass  of  the  body 
is  located  at  C  at  a  distance 
r  from  the  hinge.  A  force  P 
parallel  to  the  plane  of  the 
paper  intersects  the  plane 
through  the  hinge  and  the  center 
of  mass  at  a  distance  r'  from  0. 
The  angle  between  the  force  P 
and  the  line  OC  is  0.  It  is  de- 

sired to  find  the  hinge  reaction  when  the  body  is  rotating  about 

it  with  accelerated  angular  velocity. 
The  hinge  reaction  may  be  resolved  into  two  components. 

One  component  N  is  normal  to  OC;  the  other  component  Q 

is  in  the  direction  of  the  line  from  C  to  0. 
If  the  body  rotates  about  0  with  angular  velocity  «,    the 

centrifugal  force  which  it  exerts  on  the  hinge  is  in  the  direction 


of  OC  and  is  equal  to  -  .     The  component  of  P  in  the  direction 
of  OC  is  P  cos  (f>.     Resolving  parallel  to  CO, 


The  component  Q  depends  upon  the  angular  velocity  and  the 
applied  forces.     It  is  independent  of  the  angular  acceleration, 


CHAP.  XXIII]       ANGULAR  ACCELERATION  359 

If  the  center  of  mass  at  C  has  a  linear  acceleration  a,  the  total 
effective  force  on  the  body  in  the  direction  of  this  acceleration 

(which  is  normal  to  the  direction  of  OC)  is  —  .     Since  the  linear 

y 
acceleration  is  ra,  in  which  a  is  the  angular  acceleration  about  the 


hinge,  this  force  is  -  .     Resolving  along  an  axis  in  the  plane  of 
the  paper  normal  to  OC. 

Psin*-JV  =  ^;  (2) 

y 


N  =  P  sin  4>  -  -^-  (3) 

y 

In  Equation  (3)  the  positive  direction  of  N  is  opposite  the  direc- 
tion of  the  linear  acceleration  of  the  center  of  mass. 

The  torque  about  the  hinge  is  Prf  sin  </>.  From  Formula  XXX, 
the  angular  acceleration  is 

Tg       Pr'g  sin  <£ 

*aT      ~T- 

The  force  P  may  be  resultant  of  several  forces.  If  any  of 
the  forces  have  components  parallel  to  the  axis  of  the  hinge,  these 
components  produce  no  torque  about  the  hinge.  The  force  P 
is  the  resultant  of  all  the  components  in  planes  which  are  normal 
to  the  axis  of  the  hinge. 

Example 

A  uniform  bar  of  length  I  and  mass  ra  is  hinged  at  one  end.  Find  the 
hinge  reaction  when  the  bar  is  horizontal. 

The  force  which  produces  angular  acceleration  is  the  weight  of  m  pounds 

vertically  downward  at  the  center  of  mass  which  is  at  a  distance  of  ^  from 
the  hinge.     (In  this  example  r   =  r'  —  —) 

T  -  —     T  -  — 

'2'  3  ' 

a=lf; 

I  v  30  30 
a  =  ra  =  2  X  21  =  T 
AT  _  ma  -  _m  3g  _  m 

a  Q        4        4 


360 


MECHANICS 


[ART.   218 


When  the  bar  is  horizontal,  the  vertical  reaction  is  one-fourth  the  weight. 

To  find  the  horizontal  component  of  the  hinge  reaction,  it  is  necessary 

to  know  the  angular  velocity.     As  an  additional  condition  of  the  example, 

it  will  be  assumed  that  the  bar  was  stationary  in  the  vertical  position  and 

fell  to  the  horizontal  position 
of  Fig.  299  with  no  loss  of 
~"*\    energy.     Equating  the  work 
i    of    gravity    to    the   kinetic 
'     energy  of  rotation, 


1::, 


...1 u 

2  1 


m 


ml 
2 


T~  II 


FIG.  299. 
The  centrifugal  force  is 


ml 

ml? 

3 


ml 


3m 
"2  " 


The  force  on  the  hinge  is  to- 
ward the  center  of  mass  of  the 
bar.  The  hinge  reaction  is  hori- 
zontal toward  the  left.  The 
direction  and  magnitude  of  the 
resultant  hinge  reaction  are 
shown  in  Fig.  299,  II. 

Problems 

1.  Solve  the  Example  above 
when  the  bar  is  30  degrees 
above  the  horizontal  as  shown 
in  Fig.  300. 


m 


A  X"\  "  ** 

Ans.     Q  =  j- 


m 


FIG.  300. 


?  cos  30°  =  0.2163m. 


2.  Find  the  reactions  in  Problem  1  of  Art.  215  if  the  force  of  60  pounds  is 
vertically  downward.  Ans.     300  Ib.  vertically  upward. 

3.  Find  the  reactions  of  the  supports  for  Problem  2  of  Art.  215. 

218.  Reactions  by  Experiment. — Figure  301  shows  a  method  of 
measuring  the  reaction  of  the  bearings  which  support  a  cylinder. 
The  frame  AB  which  supports  the  bearings  is  mounted  on 
knife-edges  at  O.  The  distance  from  the  line  of  the  knife-edges 
to  the  axis  of  the  cylinder  is  equal  to  the  sum  of  the  radius 
of  the  cylinder  and  the  radius  of  the  cord  which  drives  it.  The 
line  of  the  resultant  force  in  the  cord  intersects  the  line  of  the 
knife-edges.  If  the  cylinder  is  held  so  that  it  can  not  revolve  on 
its  axis,  a  weight  hung  on  the  cord  has  no  effect  on  the  equilibrium 
of  the  frame.  If  the  cylinder  is  free  to  rotate,  there  is  an  addi- 


CHAP.  XXIII]       ANGULAR  ACCELERATION 


361 


tional  force  at  the  axis  equal  to  that  part  of  the  tension  in  the 
cord  which  goes  to  produce  acceleration.  If  there  is  friction 
at  the  bearings,  that  part  of  the  tension  in  the  cord  which  is 
required  to  overcome  the  friction  has  no  effect  on  the  balance. 
A  force  P  applied  to  the  cylinder  by  the  cord  is  transmitted  to 
the  frame  at  the  bearing.  The  moment  of  this  force  about  the 
knife-edge  at  a  distance  r  from  the  axis  is  Pr  in  a  counter-clock- 
wise direction.  If  the  cylinder  is  fixed  to  the  frame,  the  force  in 
the  cord  exerts  a  torque  Pr  upon  the  frame  at  the  bearing. 


FIG.  301. 

This  torque  is  clockwise  and  exactly  balances  the  moment  of  the 
downward  force.  Expressed  in  another  way,  the  moment  arm 
of  the  cord  about  the  line  of  the  knife-edge  is  zero.  If  the  frame 
and  cylinder  are  not  rigidly  connected,  that  part  of  the  force 
which  is  required  to  overcome  the  friction  may  be  regarded  as 
acting  on  a  rigid  body.  The  friction  at  the  bearing  exerts 
a  torque  on  the  frame  which  is  equal  and  opposite  the  moment 
of  the  corresponding  downward  force  on  the  bearing. 

To  use  the  apparatus,  the  beam  is  balanced  on  the  knife- 
edges.  The  cylinder  is  fastened  to  the  frame  so  that  it  can  not 
rotate  and  a  load  W  is  hung  on  the  cord.  If  the  knife-edges 
are  in  the  correct  position,  the  balance  is  unchanged  when  this 
load  is  applied.  The  cylinder  is  now  released  and  is  accelerated 
by  the  load.  The  part  of  the  weight  W  which  accelerates  the 
cylinder  is  transmitted  to  the  axis  of  the  bearing.  This  addi- 
tional force  downward  on  the  left  arm  of  the  frame  lifts  the  right 
arm.  The  poise  p  must  now  be  moved  toward  the  right  to 


362  MECHANICS  [ART.  219 

secure  a  running  balance.  If  the  acceleration  of  the  cylinder  is 
measured  and  if  the  moment  of  inertia  of  the  cylinder  is  known, 
the  balance  reading  makes  it  possible  to  determine  the  value 
of  g.  If  g  is  known,  the  reading  may  be  used  to  determine  the 
moment  of  inertia  of  the  cylinder.  The  results  obtained  by  this 
apparatus  require  no  correction  for  friction. 

Example 

A  cylindrical  drum  with  its  axle  weighs  8  pounds.  Its  diameter  is  a 
little  less  than  6  inches.  A  cord  is  wound  several  times  round  the  drum  and 
fastened  to  it.  The  radius  of  the  drum  plus  the  radius  of  the  cord  is  ex- 
actly 3  inches.  The  drum  is  mounted  on  a  frame  similar  to  that  of  Fig.  301. 
The  frame  is  supported  on  knife-edges  which  are  on  a  line  3  inches  from  the 
axis  of  the  axle.  The  poise  p  weighs  0.2  pound.  When  a  mass  of  1  pound 
is  hung  on  the  cord  and  the  drum  is  released,  the  poise  must  be  moved  0.96 
foot  toward  the  right  to  secure  equilibrium.  The  pound  mass  moves  28.89 
feet  in  3  seconds  after  starting  from  rest.  The  value  of  g  is  known  to  be 
32.16  feet  per  sec.  per  sec.  at  that  locality.  Find  the  moment  of  inertia  of 
the  drum. 

The  acceleration  of  the  cord  is 

2  X  28.89 

a   =  --  ;:  —  -  =  6.42  ft.  per  sec.  per  sec. 
y 

a    =  25.68  radians  per  sec.  per  sec. 
T   =  0.2  X  0.96  =  0.192ft.  Ib. 

7    =  —  ^  =  0.2404 
a 

The  force  required  to  give  an  acceleration  of  6.42  feet  per  second  per  second 
to  the  1-pound  mass  is 


The  force  required  to  produce  a  torque  of  0.192  foot-pound  on  an  arm  3 
inches  in  length  is  0.192  -=-  I  =  0.768  Ib. 

0.1996  +  0.768  =  0.9678  Ib. 
The  residue  of  0.0322  Ib.  is  taken  up  by  the  friction. 

219.  The  Reactions  of  a  System.  —  Figure  302  shows  a  frame 
which  carries  two  pulleys.  The  frame  is  supported  by  two 
knife-edges.  The  line  of  these  knife-edges  passes  through  the 
center  line  of  the  cord  which  drives  the  pulleys.  The  cord  is 
passed  over  pulley  A  and  is  wound  round  pulley  B  and  fastened 
to  it.  If  the  frame  is  in  equilibrium  with  no  load  on  the  cord,  it 
is  in  equilibrium  when  the  cord  is  loaded  and  pulley  B  is  held  by 
a  brake  or  stop  attached  to  the  frame.  The  frame  and  the  two 
pulleys  may  be  regarded  as  a  single  body  in  equilibrium.  The 


CHAP.  XXIII]      ANGULAR  ACCELERATION 


363 


moment  arm  of  a  force  applied  by  the  cord  is  zero.     The  load 
on  the  cord,  therefore,  makes  no  change  in  the  balance. 

The  frame  and  pulley  B  may  be  regarded  as  a  rigid  body, 
and  pulley  A  may  be  treated  separately.  The  simplest  form 
of  this  apparatus  is  the  one  where  the  pulleys  are  of  equal  radii  r. 


FIG.  302. 

If  pulley  A  is  frictionless,  the  pull  in  the  horizontal  cord  is  equal 
to  the  weight  W.  The  moment  of  this  pull  on  pulley  B  about 
the  knife-edges  is  equal  to  Wr.  This  moment  is  counter-clock- 
wise. The  resultant  of  the  vertical  pull  W  and  the  horizontal 
pull  W  acting  on  pulley  A  is  a  force  of  2W  sin  45°  through  the 
axis  of  the  pulley.  .  The  vertical  component  of  this  force  is  a 
force  of  2W  sin  45°  cos  45°  or  W  pounds.  The  moment  of  this 


vertical  component  about  the  line  of  the  knife-edges  is  Wr  foot- 
pounds clockwise.  The  horizontal  component  has  zero  moment 
arm.  The  total  moment  is  Wr  —  Wr  =  0. 

The  pulley  B  may  now  be  released  and  allowed  to  accelerate. 
The  friction  on  either  pulley  has  no  effect  on  the  equilibrium  of 
the  frame,  for,  in  so  far  as  friction  is  concerned,  trhe  pulleys  and 
the  frame  may  be  regarded  as  a  single  rigid  system. 


364 


MECHANICS 


[ART.   219 


If  P  is  the  force  required  to  accelerate  pulley  A,  there  is  a 
vertical  force  of  P  pounds  downward  from  the  pulley  to  the 
bearing.  This  force  produces  a  clockwise  moment  Pr  about  the 
knife-edges,  Fig.  303.  If  Q  is  the  force  which  accelerates  pulley 
B}  a  horizontal  force  equal  to  Q  may  be  regarded  as  transferred 
from  the  top  of  pulley  B  to  its  axis,  and  the  counter-clockwise 


FIG.  304. 

moment  of  the  system  is  reduced  by  an  amount  Qr.  The  total 
counter-clockwise  moment  is'  reduced  by  an  amount  (P  +  Q)r 
when  the  pulleys  are  released  and  allowed  to  accelerate.  If  the 
two  pulleys  have  equal  moments  of  inertia,  the  force  P  is  equal  to 
the  force  Q  and  the  total  change  in  moment  is  2Pr.  To  secure 
equilibrium,  the  poise  p  must  be  moved  toward  the  left. 

In  Fig.  304,  the  cord  passes  around  pulley  B  in  the  counter- 
clockwise direction.     When  this  pulley  is  accelerated,  the  force 

Q  which  produces  the  acceleration 
may  be  regarded  as  shifted  from 
the  tangent  to  the  axis.  The 
change  of  moment  is  Qr.  The 
clockwise  moment  on  the  system 
is  reduced  by  the  amount  Qr. 
When  the  moment  of  inertia  is 
the  same  for  both  pulleys,  the 
counter-clockwise  change  in  mo- 
ment which  is  due  to  the  ac- 
celeration of  pulley  B  exactly 
balances  the  clockwise  change  which  is  due  to  the  acceleration 
of  pulley  A.  There  is,  therefore,  no  change  of  the  balance  of 
the  frame,  when  the  pulleys  are  of  equal  radius  and  equal 
moment  of  inertia,  and  rotate  in  opposite  directions. 

The  discussion  of  the  preceding  paragraph  may  be  stated  in 


FIG.  305. 


CHAP.  XXIII]       ANGULAR  ACCELERATION  365 

another  way.  In  Fig.  305,  the  pulley  A  may  be  considered 
separately.  The  distance  between  the  axes  of  the  pulleys  is 
d  and  the  angle  between  the  cord  which  joins  the  pulleys  and 
the  line  which  joins  the  axes  is  6.  When  pulley  B  is  accelerated 
by  an  effective  force  Q,  the  counter-clockwise  moment  about  the 

knife-edge  is  Q  sin  0(d  +  T).     Since  sin  6  =  -p 

(1) 

The  resultant  of  two  forces  Q  acting  tangent  to  pulley  A  at  an 
angle  0  with  each  other  is 

R  =  2Q  cos  |.  (2) 

z 

This  resultant  passes  through  the  axis  of  pulley  A.  Its  vertical 
component  is 


V  =  2Qcos2  1  =  Q(l  +  cos  4>)  =  Q(l  +  sin  0)  =  Q(I  +      )  (3) 


The  moment  of  this  vertical  force  with  respect  to  the  knife-edge 
is 

M  =  Qr  +  ^.  (4) 

The  total  moment  is  the  sum  of  the  counter-clockwise  moment 
of  Equation  (1)  and  the  clockwise  moment  of  Equation   (4). 

M.  =  Qr  (5) 

in  the  counter-clockwise  direction.  Equation  (5)  states  that 
the  change  of  moment  on  the  frame  when  pulley  B  is  accelerated 
is  equal  to  the  moment  of  the  force  which  causes  the  acceleration. 
Figure  306  shows  the  author's  modification  of  the  Atwood 
machine.  A  frame  supported  by  knife-edges  carries  two  equal 
pulleys.  A  cord  which  carries  a  mass  m  runs  over  pulley  B 
and  under  pulley  A.  From  pulley  A  the  cord  runs  upward  to 
a  third  pulley  C  which  is  carried  on  a  separate  support.  The 
center  line  of  this  vertical  cord  intersects  the  line  of  the  knife- 
edges  which  carry  the  frame.  A  mass  W,  which  is  larger  than 
m  is  hung  on  the  part  of  the  cord  which  is  suspended  from  C. 
(In  Fig.  306  a  fourth  pulley  D  is  shown,  for  clearness.  A  single 
pulley  C  is  sufficient  if  placed  with  its  axis  parallel  to  the  frame. 


366 


MECHANICS 


[ART.   220 


A  brake  (not  shown  in  the  drawing)  holds  pulley  C  stationary 
while  the  poise  is  adjusted  to  balance.  When  the  brake  is 
released,  the  mass  m  is  accelerated  upward.  The  additional 
force  on  the  cord  which  supports  the  mass  m  is 


p  = 

Q 


(6) 


This  additional  force  is  balanced  by  moving  the  poise  p  toward 
the  left.  A  few  trials  are  required  to  find  the  position  of  running 
equilibrium. 


FIG.  306. 


If  I  is  the  moment  arm  of  the  mass  m,  p  is  the  mass  of  the 
poise,  and  b  is  the  distance  the  poise  is  moved  to  secure  equi- 
librium of  the  frame, 

ma 
Q 


(7) 


This  apparatus  is  free  from  any  correction  for  friction  or  the 
mass  of  the  pulleys.  To  avoid  any  correction  for  the  weight 
of  the  cord,  it  should  be  continuous  from  W  to  m.  The  mass 
m  includes  the  mass  of  cord  above  and  below  it. 

220.  Summary. — Angular  acceleration  bears  the  same  rela- 
tion to  linear  acceleration  that  angular  velocity  bears  to  linear 
velocity. 

_  do>  _  (P8 
1  ~  dt    '"  dt2' 


Angular  acceleration 


CHAP.  XXIII]       ANGULAR  ACCELERATION  367 

Angular    acceleration    is    proportional    to    the    torque    and 
inversely  proportional  to  the  moment  of  inertia  of  the  body. 
Moment  of  inertia  X  angular  acceleration 


Torque 


9 


T  =  —•  Formula  XXX 

9 

When  the  foot  is  the  unit  of  length  in  the  computation  of  torque 
and  moment  of  inertia,  the  value  of  g  is  32.174. 

The  component  of  the  reaction  at  the  axis  in  the  direction 
normal  to  the  plane  through  the  axis  and  the  center  of  mass  is 

ma 
N  =  P  cos  6 , 

9 

in  which  P  cos  6  is  the  component  of  the  applied  force  normal 
to  the  plane  through  the  axis  and  the  center  of  mass,  m  is  the 
mass,  and  a  is  the  linear  acceleration  of  the  center  of  mass. 
The  component  N  is  positive  in  the  direction  opposite  the  direc- 
tion of  the  linear  acceleration  of  the  center  of  mass.  The  linear 
acceleration  a  is  given  by  the  equation 

a  =  ra, 

in  which  r  is  the  distance  of  the  center  of  mass  from  the  axis  of 
rotation. 

The  component  of  the  reaction  along  the  line,  perpendicular 
to  the  axis,  which  joins  the  center  of  mass  with  the  axis  of  rotation 
is  equal  to  the  component  of  the  applied  forces  in  this  direction 
combined  with  the  centrifugal  force.  If  the  axis  of  rotation 
passes  through  the  center  of  mass,  the  centrifugal  force  is  zero 
and  the  linear  acceleration  of  the  center  of  mass  is  zero;  the 
reaction  of  the  axis  is  equal  and  opposite  to  the  applied  force. 

In  the  calculation  of  the  acceleration  of  a  rotating  body, 
the  actual  mass  may  be  replaced  by  an  equivalent  mass  in  the 
line  of  action  of  the  applied  force.  If  r'  is  the  perpendicular 
distance  from  the  axis  to  the  line  of  action  of  the  applied  force, 
7  is  the  moment  of  inertia  of  the  body,  and  m'  is  the  equivalent 
mass, 

7  =  m'r'2; 
7 


'2 


By  means  of  the  equivalent  mass,  a  problem  of  angular  accelera- 
tion may  be  reduced  to  a  problem  of  linear  acceleration.  The 
angular  velocity  and  acceleration  may  finally  be  computed 
from  the  linear  acceleration  and  velocity  by  dividing  by  r'. 


CHAPTER  XXIV 
ANGULAR  VIBRATION 

221.  Work  of  Torque.  —  When  a  force  is  applied  to  an  arm  of 
length  r,  the  work  done  by  the  force  when  the  arm  turns  through 
an  angle  6  is 

U  =  Pr6,  (1) 

in  which  P  is  the  component  of  the  force  normal  to  the  plane 
through  its  point  of  application  and  the 
axis  of  rotation.  Since  the  product  P  X  r 
is  the  torque,  Equation  (1)  is  equivalent  to 

u  =  Te.  (2) 

Equation  (2)  states  that  the  work  of  a 
constant  force  is  equal  to  the  product  of 
the  torque  multiplied  by  the  angular  dis- 
placement. The  work  per  revolution  is 

U  =  2vT.  (3) 

When  the  torque  varies, 

U  =  fTde  (4) 

When  an  elastic  rod,  or  a  spiral  or  helical  spring  is  twisted, 
the  torque  is  proportional  to  the  angular  displacement.  If  K 
is  the  torque  when  the  angular  twist  is  one  radian,  the  expression 
for  the  torque  when  the  angle  is  6  radians  is 

T  =  KB.  (5) 

u  =  Kfede  =  ![*']£  -  f  M  -  of).  (6) 

When  the  initial  angular  displacement  is  zero 

(7) 


Problems 

1.  A  force  of  20  pounds  at  the  end  of  an  arm  4  feet  in  length  ft  fists  a 
steel  rod  through  an  angle  of  90  degrees.  If  there  is  no  initial  torque,  find 
the  work,  Solve  by  means  of  the  average  torque. 

Ans.     U  -  ?2+_?  x  5  -  62.832  ft,  lb. 

_  & 


CHAP.  XXIV]  ANGULAR  VIBRATION  369 

2.  A  wheel  is  driven  by  a  constant  torque.     Find  the  work  per  revolution. 

Ans.     U  =  2irT. 

3.  A  drum  16  inches  in  diameter  is  driven  by  a  rope  which  is  wound 
round  it.     The  rope  is  £  inch  in  diameter  and  exerts  a  pull  of  300  pounds. 
Find  the  work  per  revolution. 

Ans.     U  =  1295.9  ft.  Ib. 

222.  Angular  Velocity  from  Variable  Torque. — When  the 
torque  is  constant,  the  angular  velocity  is  easily  obtained  first 
by  finding  the  angular  acceleration  and  then  by  multiplying  by 
the  time.  When  the  torque  is  variable,  the  angular  velocity  at 
any  given  position  is  easiest  found  by  means  of  the  kinetic  energy 
of  rotation. 

The  most  important  problem  is  that  in  which  the  torque 
varies  as  the  angular  displacement.  If  a  body  on  an  elastic 
rod  or  spring  is  twisted  through  an  angle  |8  from  the  position  of 

Jf  f)2 

equilibrium,  the  work  done  is  — ~ —     This  work  is  stored  up  as 

m 

elastic  energy.  If  the  body  is  released  and  permitted  to  rotate 
back  to  the  position  at  which  its  displacement  from  the  position 
of  equilibrium  is  6,  the  change  in  elastic  energy  is  the  first  member 
of  the  equation 

KP       K62       7o>2 

-~ -  =  — •  Formula  XXXI 


The  second  member  of  Formula  XXXI  is  the  kinetic  energy  of 
rotation.  The  formula  states  that  the  change  of  potential  en- 
ergy is  equal  to  the  kinetic  energy,  or  that  the  sum  of  the  potential 
energy  and  the  kinetic  energy  is  constant.  The  angle  0  is 
the  amplitude  of  the  vibration  in  angular  measure. 
Formula  XXXI  reduces  to 

"  =-?2.  (i) 


The  maximum  angular  velocity  is 

(2) 


Problems 

1.  A  homogeneous  solid  cylinder   1  foot  in  diameter  and  weighing  20 
pounds  is  suspended  with  its  axis  vertical  by  means  of  a  |-inch  steel  rod  10 

24 


370 


MECHANICS 


[ART.   223 


feet  in  length,  Fig.  308.     A  force  of  3  pounds  at  the  end  of  an  arm  1  foot 
long  is  capable  of  twisting  this  rod  through  one  radian.     The  cylinder  is 


FIG.  308. 

rotated  through  an  angle  of  180  degrees  and  then  released.     Find  its  maxi- 
mum angular  velocity. 


Ans.     w2  =  ^-',  w  =  19.52  radians  per  sec. 

A.O 

2.  In  problem  1,  what  is  the  maximum  angular  velocity  if  the  initial 
displacement  is  90  degrees  from  the  position  of  equilibrium? 

223.  Time  of  Vibration.— Formula  XXXI 
for  the  angular  velocity  is  similar  to  For- 
mula XXVI  for  the  linear  velocity  caused  by 
a  force  which  varies  as  the  displacement. 
In  Formula  XXXI,  angular  velocity  and 
angular  displacement  take  the  place  of 
linear  velocity  and  displacement,  and 

_ ^  moment  of  inertia  replaces  mass. 

a  t>  c  ct  «       if  £ne  circular  motion  of  a  point  on  a 

body  which  moves  with  the  angular  velocity 

of  Formula  XXXI  is  developed  into  linear  motion,  as  shown 

in  Fig.  309,  this  developed  motion  is  simple  harmonic.     If  a 

sheet  of  paper  is  placed  on  a  cylinder  which  has  this  motion, 


CHAP.   XXIV] 


ANGULAR  VIBRATION 


371 


and  a  pencil  is  moved  with  uniform  speed  parallel  to  its  length, 
the  pencil  mark  is  the  sine  curve  shown  in  Fig.  310. 

Since  the  angular  motion  of  a  body  subjected  to  torque  which 
varies  as  the  angular  displacement  is  a  simple  harmonic  motion, 
it  may  be  studied  by  means  of  a  circle  of  reference.  The  radius 
of  this  circle  of  reference  is  the  angular  amplitude  /3  expressed 
in  radians,  Fig.  311.  A  point  may  be  con- 
sidered as  moving  in  the  circumference  of 
this  circle  with  a  linear  velocity  which  is 
equal  to  the  maximum  angular  velocity  of  the 
vibrating  mass.  The  time  of  a  complete  period 


FIG.  311. 


is  the  time  required  for  this  point  to  make  one  complete  revolu- 
tion around  the  circle  of  reference.  The  angular  velocity  at  any 
instant  is  the  component  along  the  diameter  of  the  circle  of 
this  maximum  velocity  in  its  circumference. 


Problems 

1.  Find  the  time  of  a  complete  oscillation  of  the  cylinder  of  Problem  1 
of  Art.  222. 

Ans.     tc  =  —  =  2ir\~  =  1.01  sec. 

CO  \   30 

2.  Find  the  time  of  a  complete  oscillation  for  Problem  2  of  Art.  222. 

3.  Express  in  terms  of  7  and  K  the  time  of  a  complete  oscillation  caused 
by  a  torque  which  varies  as  the  displacement.     Compare  the  result  with 
Equation  (10)  of  Art.  185. 


Formula  XXXII 


Ans.     tc  =  2ir\ 


4.  A  cylindrical  disk  6  inches  in  diameter  is  suspended  with  its  axis  vertical 
by  means  of  a  vertical  wire.  The  disk  makes  100  complete  oscillations  in 
96  seconds.  Two  wires,  each  0.04  inch  in  diameter,  are  fastened  to  the 
cylinder  and  extend  horizontally  to  smooth  pulleys,  as  shown  in  Fig.  308. 


372 


MECHANICS 


[ART.   223 


When  a  mass  of  1  pound  is  hung  on  each  wire,  the  cylinder  is  found  to 
rotate  12  degrees.     Find  the  moment  of  inertia  of  the  cylinder. 


2  X  3.02 


K  = 


3.02 


7T 

15 


tc  = 


4?r 


15.10, 

27T 


pound 


B,  12  of/am 


FIG.  312. 


Log  /   =  0.25649,  7    =   1.805 
and  foot  units. 

6.  The  small  cylinder  .4  of  Fig.  312  sus- 
pended by  a  wire  is  found  to  make  1000 
complete  torsional  vibrations  in  240 
seconds.  A  bronze  disk  B,  12  inches 
outside  diameter,  1  inch  inside  diameter, 
and  weighing  32.04  pounds  is  hung  on  the 
small  cylinder.  The  time  of  1000  complete 
vibrations  is  now  2500  seconds.  Find  the 
moment  of  inertia  of  the  small  cylinder. 


Ans. 


I  of  disk  =  16.02  X 
0.0576 


145 


4.0329; 


0.242 
2.52 


6.25  -  0.0576 


576  '   /o  +  . 

;  Log  /o  =  2.57417;  /„  =  0.0375. 


6.  Two  wires,  each  0.06  inch  in  diameter  are  attached  to  the  disk  of 
Problem  5  and  passed  over  smooth  pulleys.  When  a  1-pound  mass  is  hung 
on  each  wire,  the  disk  is  twisted  through  an  angle  of  45°  50'.  Find  g. 


I*" 


T 


Front  View 

FIG.  313. 


End  View 


7.  The  frame  of  Fig.  313  is  supported  by  a  vertical  wire  at  the  center  of 
the  vertical  cylinder  A.  Each  of  the  cylinders  B  and  C  weighs  3  pounds 
and  has  its  center  of  gravity  2  inches  from  the  end  nearest  to  cylinder  A. 
When  the  inner  ends  of  B  and  C  are  0.250  inch  from  the  vertical  axis,  the 
time  of  vibration  of  the  system  is  0.428  second.  When  cylinders  B  and  C 


CHAP.   XXIV] 


ANGULAR  VIBRATION 


373 


are  moved  out  till  their  ends  are  10  inches  from  the  vertical  axis,  the  time 
of  vibration  is  1.126  second.  Find  the  effective  moment  of  inertia  of  the 
system  with  the  cylinders  in  the  first  position. 

Ans.     1 2  -  1 1  =  5.7891;  72  =  6.7667;  A  =  0.9776, 

in  which  72  is  the  moment  of  inertia  at  the  second  position  and  I\  is  the 
moment  of  inertia  at  the  first  position. 

8.  A  torque  of  3.436  foot  pounds  twists  the  system  of  Problem  7  through 
an  angle  of  30  degrees.  Find  g. 

The  method  of  Problems  5  and  6  may  be  used  to  determine  the  value 
of  g  at  any  locality.  The  mass  of  the  cylinder  may  be  determined  by 
accurate  weighing.  It  is  not  necessary  that  standard  weights  be  used  in 
this  weighing.  It  is  necessary,  however,  that  the  weights  used  in  weighing 
the  disk  and  those  used  in  determining  the  torque  be  expressed  in  terms  of 
the  same  unit.  This  method  assumes  that  the  material  of  the  disk  is 
homogeneous  throughout. 

The  method  of  Problems  7  and  8  does  not  require  that  the  cylinders  B 
and  C  should  be  homogeneous.  The  method,  however,  does  require  that 
the  location  of  their  centers  of  gravity  be  experimentally  determined. 

224.  The  Gravity  Pendulum. — A  body  which  is  free  to  rotate 
through  a  small  angle  about  a  horizontal  axis  under  the  action 
of  its  weight  is  called  a  pendulum.  The  axis  0 
of  Fig.  314  about  which  it  rotates  is  called  the 
axis  of  suspension.  If  m  is  the  mass  of  the 
pendulum  and  f  is  the  distance  of  its  center  of 
mass  below  the  axis  of  suspension,  the  torque 
which  is  due  to  its  weight  .when  the  pendulum 
is  displaced  through  an  angle  6  from  the  position 
of  equilibrium  is  given  by  the  equation 

T  =  mr  sin0    •  (1) 

If  0  is  small,  its  sine  is  practically  equal  to  its 
arc  in  radian  measure.  For  a  small  angle, 
Equation  (1)  becomes 

T  =  mrB.  (2) 


FIG.  314. 


When  the  angular  displacement  is  small,  the  torque  is  propor- 
tional to  the  displacement  and  Formula  XXXII  applies  to  the 
motion  of  the  pendulum.  The  value  of  K  for  a  pendulum 
vibrating  through  a  small  arc  is  K  =  mr,  and  Formula  XXXII 
becomes, 

(3) 


mrg 


374 


MECHANICS 


[ART.   225 


Problems 

1.  A  uniform  bar  of  length  I  vibrates  as  a  pendulum  about  an  axis  per- 
pendicular to  its  length  through  one  end.  Find  the  time  of  a  complete 
period. 


2.  Solve  Problem  1  for  a  length  of  4  feet. 


AnS.      tc    =    27T\    TT- 


Ans.     tc  =  1.809  sec. 


3.  A  flywheel  is  4  feet  in  diameter, 
weighs  200  pounds,  and  has  its  center 
of  mass  at  its  axis.  The  flywheel  vi- 
brates as  a  pendulum  about  a  knife 
edge  under  the  rim  at  a  distance  of  23 
inches  from  the  axis  (Fig.  315).  The 
wheel  makes  100  complete  vibrations  in 
196  seconds.  Find  its  moment  of  inertia 
with  respect  to  the  knife  edge  and  its 
moment  of  inertia  with  respect  to  its  axis. 
Ans.  I  =  1200.2;  70=  465.5. 

225.  Simple  Pendulum.—  The 
ideal  simple  pendulum  consists  of 
a  small  body  suspended  by  a 
weightless  cord.  The  dimensions 
of  the  body  must  be  so  small  that  all  parts  are  at  the  same  dis- 
tance from  the  axis  of  suspension.  If  m  is  the  mass  of  the  body 
and  I  is  its  distance  from  the  axis  of  suspension, 

r  =  l;I  =  ml2, 
and  Equation  (3)  of  the  preceding  article  becomes 


FIG.  315. 


Problems 

1.  Find  the  length  of  the  simple  pendulum  which  makes  a  single  oscilla- 
tion in  one  second  at  a  place  where  g  =  32.174  ft. 

Ans.     I  =  39.12  in. 

2.  Find  the  value  of  g  at  a  place  where  the  "second  pendulum"  is  39.00 
inches  in  length. 

In  the  ordinary  physical  pendulum, 


=  m(r2  -f-  kl),  (2) 

in  which  k0  is  the  radius  of  gyration  with  respect  to  the  axis 
parallel  to  the  axis  of  suspension  which  passes  through  the  center 


CHAP.  XXIV]  ANGULAR  VIBRATION  375 

of  mass.     The  expression  for  the  time  of  a  complete  period  is 

*•"*     Ir  +  T  .  (3) 

VT 

/Co 

The  term  r.+  -=^  in  Equation  (3)  is  a  length  and  corresponds 

with  I  of  Equation  (1).     This  term  is  called  the  length  of  the 
equivalent  simple  pendulum.     It  will  be  represented  by  /'. 

r«f  +  y-  (4) 

If  k0  is  small  compared  with  r}  then  Z'  approaches  r,  and  the 
pendulum  is  approximately  a  simple  pendulum. 

Problem 

3.  A  sphere  0.4  inch  in  diameter  is  suspended  by  a  cord  of  negligible 
weight.  The  center  of  the  sphere  is  40  inches  from  the  point  of  suspension 
of  the  cord.  What  is  the  length  of  the  equivalent  simple  pendulum?  What 
is  the  relative  error  in  the  time  of  vibration  if  the  radius  of  gyration  of  the 
sphere  is  neglected? 

Ana.     I'  =  40.004  in.;  error  1  part  in  20,000. 

226.  Axis  of  Oscillation. — The  axis  of  oscillation  of  a  physical 
pendulum  is  a  line  parallel  to  the  axis  of  suspension  at  a  distance 
from  the  axis  of  suspension  equal  to  the  length  of  the  equivalent 
simple  pendulum.  The  axis  of  oscillation  lies  in  the  plane  which 
passes  through  the  axis  of  suspension  and  the  center  of  mass  of 
the  pendulum.  A  particle  at  the  axis  of  oscillation  is  neither 
accelerated  nor  retarded  by  the  motion  of  the  pendulum  as  a 
whole.  A  particle  below  the  axis  of  oscillation  is  forced  to  vibrate 
faster  than  it  would  if  it  were  suspended  by  a  weightless  cord  and 
moved  as  a  simple  pendulum.  A  particle  above  the  axis  of  oscil- 
lation moves  slower  than  it  would  if  it  were  moving  alone. 

If  V  is  the  distance  between  the  axis  of  suspension  and  the 
axis  of  oscillation, 

r  r 

Problems 

1.  A  bar  of  uniform  section  and  density  vibrates  about  an  axis  which  is 
perpendicular  to  its  length  and  passes  through  one  end.  Find  the  distance 
between  the  axis  of  suspension  and  the  axis  of  oscillation. 

Ans.     I'  =  —• 


376  MECHANICS  [ART.   227 

2.  A  uniform  bar  of  length  I  vibrates  as  a  pendulum  about  an  axis  per- 
pendicular to  its  length  at  a  distance  of  one-third  its  length  from  one  end. 
Find  the  length  of  the  equivalent  simple  pendulum. 

„       2Z 

Ans.     I    —  —  • 
o 

3.  Find  the  length  of  the  simple  pendulum  equivalent  to  a  uniform  bar 
of  length  I  which  vibrates  about  an  axis  of  suspension  at  one-fourth  the 
length  from  one  end.     Solve  also  when  the  axis  of  suspension  is   three- 
eighths  the  length  from  one  end. 

Ans     ?  -7l  .  v  -19J 
12  J  '     '   24  ' 

4.  What  is  the  position  of  the  axis  of  suspension  of  a  uniform  bar  which 
gives  the  minimum  time  of  vibration?     What  is  the  length  of  the  equivalent 
simple  pendulum? 

Ans.      r  =  -4=-;  V  =  -7=-  =  °-577  L 
Vl2  V3 

227.  Exchange  of  Axes.  —  If  the  axis  of  oscillation  of  a  pendu- 
lum is  made  the  axis  of  suspension,  the  former  axis  of  suspen- 
sion becomes  the  axis  of  oscillation.  When  the  original  axis 
of  suspension  is  at  a  distance  r  from  the  center  of  gravity  of  the 
pendulum,  the  length  of  the  equivalent  simple  pendulum  is 

^2 

I'  -  r  +  =?-  (1) 


The  distance  from  the  center  of  gravity  to  the  axis  of  oscilla- 

fc2 

tion  is  -^-     When  the  axis  of  oscillation   becomes  the  axis  of 
r 

k2 

suspension,  -=  is  the  distance  from  the  new  axis  of  suspension  to 
r 

the  center  of  gravity.  If  I"  is  the  length  of  the  equivalent  simple 
pendulum  when  the  axis  of  oscillation  is  made  the  axis  of  sus- 
pension, 

fr2  ^2  JL2 

I"  -=?  +  §•*  3  +  r,  (2) 

r        «2        r 

r 
I"  =  V.  (3) 

The  principle  of  Equation  (3)  is  employed  in  the  determination 
of  g.  A  pendulum  is  provided  with  a  pair  of  knife-edges  parallel 
to  each  other  at  approximately  the  correct  distance  for  one 


CHAP.  XXIV]  ANGULAR  VIBRATION  377 

knife-edge  to  be  the  axis  of  suspension  and  the  other  the  axis  of 
oscillation.  A  small  body,  b  of  Fig.  316,  is  arranged  to  slide 
along  the  pendulum.  This  body  is  adjusted  till  the  time  of 
vibration  is  the  same  when  either  axis  is  used  as  the  axis  of 
suspension.  The  time  of  vibration  and  the  distance 
between  the  knife-edges  afford  the  necessary  data  for 
the  determination  of  g  by  substitution  in  the  equa- 
tion of  vibration  of  the  simple  pendulum. 

Instead  of  one  adjustment  weight,  two  equal 
weights  may  be  employed.  These  may  be  moved 
equal  distances  in  opposite  directions  at  each  adjust- 
ment. When  the  adjustment  is  made  in  this  way,  r 
is  not  changed,  the  variation  of  time  of  vibration 
follows  a  simpler  law,  and  correct  position  is  easier  to 
find. 

The  determination  of  g  by  this  method  involves  only  measure- 
ments of  time  and  length,  and  is  independent  of  mass  and  density. 

228.  Summary.  —  When    the    torque    varies    as  the    angular 
displacement,   the  work  is  given  by 

V  -  f  (S\  -  9J), 

in  which  K  is  the  torque  when  the  displacement  from  the  posi- 
tion of  zero  torque  is  one  radian,  and  6\  and  62  are  angular  dis- 
placements from  the  position  of  equilibrium.  When  the  initial 
displacement  is  zero  and  the  final  displacement  is  0  radians, 

TJ  -  K6* 

~2~ 

The  equation  which  connects  the  work  of  torque  and  the  kinetic 
energy  is 


7co2  _,          , 

—  =  -—}  Formula 


Z  Z  Zg 

in  which  /?  is  the  angular  displacement  when  the  velocity  is  zero, 
and  co  is  the  angular  velocity  when  the  displacement  is  6.  When 
6  is  zero 


If  angular  displacement  is  plotted  on  a  straight  line,   the 
motion  of  a  body  subjected  to  torque  which  varies  as  the  dis- 


378  MECHANICS  [ART.   228 

placement  is  a  simple  harmonic  motion.     The  time  of  a  single 
oscillation  is 


The  time  of  a  complete  period  is 

Formula  XXXII 


In  a  gravity  pendulum  the  torque  is 
T  =  mr  sin  0. 

When  the  angle  is  small,  sin  6  reduces  to  6  and 

T  =  mr0. 

With  a  small  amplitude  of  vibration  the  torque  on  a  gravity 
pendulum  varies  as  the  displacement  and  Formula  XXXI 
applies.  The  constant  K  is  equal  to  mr,  and  the  time  of  a 
complete  period  is 


tc  =  2ir 


The  term  r  +  ~=  is  the  length  of  the  equivalent  simple  pen- 

dulum. The  length  of  the  equivalent  simple  pendulum  is  the 
distance  between  the  axis-  of  suspension  and  the  axis  of  oscillation. 
The  axis  of  suspension  and  the  axis  of  oscillation  are  inter- 
changeable. If  the  time  of  vibration  is  found  for  a  given  axis 
of  suspension  and  the  pendulum  is  then  reversed  and  suspended 
from  the  axis  of  oscillation,  the  time  of  vibration  for  the  two 
positions  is  found  to  be  identical. 


CHAPTER  XXV 
MOMENTUM  AND  IMPULSE 

229.  Momentum.  —  The  product  of  the  mass  of  a  body  multi- 
plied by  its  velocity  is  called  its  momentum. 

Momentum  =  mv.  (1) 

Since  velocity  is  a  vector  quantity  and  mass  is  a  scalar  quan- 
tity, their  product  is  a  vector.  Momentum  has  direction  as  well 
as  magnitude. 

Problems 

1.  What  is  the  total  momentum  of  40  pounds  moving  east  50  feet  per 
second  and  20  pounds  moving  west  60  feet  per  second? 

Ans.     Momentum  =  800  in  foot  and  pound  units. 

2.  What  is  the  total  momentum  of  20  pounds  moving  north  30  feet  per 
second  and  40  pounds  moving  east  20  feet  per  second? 

An  s  .     Momentum  =  1000  units  north  53°  08'  east. 

230.  Impulse.  —  The  rate  of  change  of  momentum  when  the 

velocity  changes  is  ra-^  which  is  equal  to  ma. 

dv 
m-r- 

-^  ma          dt  ,.,,. 

Force  =  —  =  -  >  (1) 

9          9 


Pdt  =         ;  (2) 

;  (3) 


f 
J 


Pdt  =  -(v2  -  Vl).  (4) 


The  term  J.  Pdt  is  called  the  impulse  of  the  force.  When  the 
force  is  constant  during  an  interval  of  time  t,  the  impulse  is 
Pt.  Equation  (4)  shows  that  the  change  of  momentum  is 
proportional  to  the  impulse.  When  the  pound  force  is  used  as 
the  unit  of  force  and  the  pound  mass  is  used  as  the  unit  of  mass, 
the  impulse  is  the  change  in  momentum  divided  by  32.174. 
In  terms  of  the  absolute  units,  the  change  of  momentum  is 
numerically  equal  to  the  impulse. 

379 


380  MECHANICS  [ART.  231 

When  an  impulse  is  given  by  a  blow,  as  when  an  object  is 
struck  by  a  hammer,  the  force  is  large  and  the  time  is  very  short. 
Such  an  impulse  is  frequently  called  an  instantaneous  impulse. 
Sudden  impulse  is  a  better  name.  Suppose  that  a  ball  moving 
with  a  velocity  of  100  feet  per  second  strikes  a  wall  and  bounds 
back  with  a  velocity  of  nearly  100  feet  per  second.  Suppose  that 
the  compression  of  the  ball  amounts  to  0.5  inch.  If  the  force 
were  uniform  during  the  time  of  contact,  the  average  velocity 
would  be  50  feet  per  second  while  the  ball  is  coming  to  rest. 
The  time  required  to  bring  the  ball  to  rest  would  be 

•-        =  second. 


The  entire  time  of  contact  would  be  about  -$$-$  second.  If 
the  pressure  varies  as  the  displacement,  instead  of  being  uniform, 
the  motion  during  contact  is  simple  harmonic.  The  time  re- 
quired to  complete  a  semicircle  of  \  inch  radius  at  a  speed  of 

100  feet  per  second  is  ?>TQQ  =  0.0013  second.     When  a  hammer 

strikes  an  anvil,  the  deformation  is  much  smaller  and  the  time  of 
contact  is  less.  In  general,  the  duration  of  a  so-called  instanta- 
neous impulse  is  a  few  ten-thousandths  of  a  second. 

It  is  sometimes  stated  that  an  instantaneous  impulse  is  com- 
pleted before  the  body  which  is  struck  starts  to  move.  This 
statement  is  incorrect.  The  full  velocity  is  attained  at  the 
end  of  the  impulse.  The  time,  however,  is  so  short  that  the 
actual  displacement  is  small.  In  the  case  of  a  galvanometer 
needle,  which  is  actuated  by  the  discharge  of  a  condenser,  the 
needle  has  attained  its  maximum  velocity  at  the  time  when  the 
current  becomes  negligible.  The  deflection,  on  the  other  hand, 
is  so  small  that  the  moment  arm  of  the  impulse  is  practically 
constant  throughout  the  entire  impulse. 

231.  Action  and  Reaction.  —  When  force  acts  on  a  body,  the 
force  must  come  from  some  other  body.  Since,  as  stated  by 
Newton's  Third  Law,  action  and  reaction  are  equal,  the  force 
exerted  by  the  second  body  on  the  first  body  is  equal  and  opposite 
the  force  exerted  by  the  first  body  on  the  second  body.  Since 
the  forces  are  equal,  the  impulses  are  equal,  and,  conse- 
quently, the  change  of  momentum  of  one  body  is  equal  and  oppo- 
site the  change  of  momentum  of  the  other. 

In  Fig.  317,  A  and  B  are  two  bodies.  Suppose  that  A  exerts 
a  force  on  B  by  means  of  a  spring  which  forms  a  part  of  A  .  The 


CHAP.  XXV]       MOMENTUM  AND  IMPULSE  381 

change  of  momentum  of  A  toward  the  left  is  equal  and  opposite 
the  change  of  momentum  of  B  toward  the  right.  Suppose  that 
A  with  the  attached  spring  weighs  5  pounds  and  B  weighs  4 
pounds,  and  suppose  that  both  are  stationary.  When  the 
spring  is  released,  suppose  that  B  receives  a  velocity  of  10  feet 
per  second  toward  the  right.  Its  change  of  momentum  is  40. 
At  the  same  time,  A  receives  a  velocity  of  8  feet  per  second 
toward  the  left.  Its  change  of  momentum 
is  ~40.  The  total  change  of  momentum 
of  the  two  bodies  considered  as  a  single 
system  is  zero.  It  is  impossible  to  change 
the  total  momentum  of  two  bodies  by  FIG.  317. 

means  of  any  force  exerted  between  them. 

Any  change  in  the  momentum  of  a  body  is  at  the  expense  of 
an  equal  and  opposite  change  in  the  momentum  of  some  other 
body  or  bodies.  When  a  man  jumps  upward  into  the  air,  he 
kicks  the  earth  in  the  opposite  direction  with  equal  momentum. 
While  he  is  in  the  air,  the  attraction  of  gravity  draws  him  toward 
the  earth  and,  likewise,  draws  the  earth  toward  him.  Con- 
sidered as  one  body,  the  center  of  gravity  of  the  earth  and  the 
man  moves  on  unchanged. 

If  mi  is  the  mass  of  one  body  and  v\  is  its  velocity,  ra2  is  the 
mass  of  a  second  body  and  v%  is  its  velocity,  and  if  a  force  acts 
between  the  bodies  which  changes  the  velocity  of  mi  from  v\ 
to  v(  and  changes  the  velocity  of  ra2  from  vz  to  v'z, 


+  m2v2  =  miv'i  +  m^.  (1) 


Example 


A  mass  of  20  pounds  moving  east  30  feet  per  second  overtakes  a  mass  of 
12  pounds  moving  east  12  feet  per  second.  After  collision,  the  bodies  move 
together  with  the  same  velocity.  Find  that  velocity. 

The  total  momentum  east  is 

20  X  30  =  600 
12  X  10  =  120 

720 
tf* 

The  total  Hf&ss  is  32  pounds.     If  v  is  the  final  velocity, 

32v  =  720 

v  =  22.5  ft.  per  sec.  east. 


382  MECHANICS  [ART.  232 

Problems 

1.  A  man  weighing  150  pounds  jumps  horizontally  with  a  speed  of  10  feet 
per  second  from  a  boat  weighing  200  pounds.     If  the  boat  was  initially  at 
rest  and  if  the  friction  of  the  water  is  neglected,  with  what  velocity  does  the 
boat  move  in  the  opposite  direction? 

Ans.     7.5  ft.  per  sec. 

2.  A  man  weighing  150  pounds  can  jump  with  a  horizontal  velocity  of  15 
feet  per  second  with  reference  to  a  fixed  point.     With  what  speed  relative 
to  the  earth  can  he  jump  from  a  boat  which  weighs  100  pounds  and  what  is 
the  velocity  of  the  boat? 

Ans.     6  ft.  per  sec.;  9  ft.  per  sec. 

3.  A  shot  weighing  10  pounds  is  fired  from  a  gun  weighing  800  pounds. 
The  velocity  of  the  shot  relative  to  the  earth  is  1200  feet  per  second.     Find 
the  velocity  of  the  gun's  recoil.     Find  the  kinetic  energy  of  the  shot  and  of 
the  gun. 

Ans.  v  =  15  ft.  per  sec.;  kinetic  energy  of  shot  =  223,780  ft.  lb.,  kinetic 
energy  of  gun  =  2797  ft.  lb. 

4.  When  a  shot  is  fired  from  a  gun,  show  that  the  kinetic  energy  of  the 
shot  is  to  the  kinetic  energy  of  the  gun  as  the  weight  of  the  gun  is  to  the 
weight  of  the  shot. 

5.  A  4-pound  mass  moving  east  60  feet  per  second  meets  a  5-pound  mass 
moving  west  20  feet  per  second.     After  collision,  the  5-pound  mass  moves 
east  20  feet  per  second.     By  means  of  the  total  momentum  find  the  velocity 
of  the  4-pound  mass. 

Ans.     v  =  10  ft.  per  sec.  east. 

6.  If  the  5-pound  mass  of  Problem  5  moves  east  40  feet  per  second  after 
collision,  what  is  the  velocity  of  the  4-pound  mass?  . 

Ans.     v  =  15  ft.  per  sec.  west. 

7.  The  bodies  of  Problem  5  collide  obliquely.     After  collision  the  5-pound 
mass  moves  northeast  with  a  velocity  of  30  feet  per  second.     What  is  the 
velocity  of  the  4-pound  mass? 

Ans.  The  4-pound  mass  has  a  velocity  component  east  of  10.97  ft.  per 
sec.  and  a  component  south  of  26.51  ft.  per  sec. 

232.  Collision  of  Inelastic  Bodies.  —  When  inelastic  bodies 
collide,  -they  remain  together  after  collision  and  move  with 
the  same  velocity.  If  mi  and  m%  are  the  masses,  v\  and  v^  are 
their  respective  velocities  before  collision,  and  v  is  their  common 
velocity  after  collision,  the  momentum  equation  is  • 

miVi  +  m2vz  =  (mi  +  m^v.  (1) 

When  inelastic  bodies  collide,  there  is  a  loss  of  kinetic  energy. 
[6  of  the  kinetic  energy  is  transformed  into  heat  energy. 


Problems 

1.  A  mass  of  20  pounds  moving  east  with  a  velocity  of  40  feet  per  second 
overtakes  a  mass  of  30  pounds  moving  east  with  a  velocity  of  10  feet  per 
second.  After  collision  the  bodies  move  together.  Find  their  common 
velocity.  Ans.  v  =  22  ft.  per  sec. 


CHAP.  XXV]       MOMENTUM  AND  IMPULSE 


383 


2.  In  Problem  1,  what  part  of  the  kinetic  energy  is  lost? 

Ans.     41  per  cent. 

3.  A  mass  of  20  pounds  moving  east  with  a  velocity  of  40  feet  per  second 
meets  a  mass  of  30  pounds  moving  west  with  a  velocity  of  10  feet  per  second. 
Find  the  velocity  after  collision  and  the  kinetic  energy  lost. 

Ans.     10  ft.  per  sec.;  87.8  per  cent. 

4.  A  gun  is  hung  in  a  horizontal  position 
on  a  vertical  support  which  is  hinged  at  the 
top  as  shown  in   Fig«   318.     The  gun  and 
support  weigh  800  pounds  and  their  center 
of  gravity  is  12  feet  below  the  hinge.     When 
a  10-pound  shot  is  fired  from  the  gun,  the 
reaction  swings  the  support  through  an  arc 
of  45  degrees.     Find  the  velocity  of  the  shot. 

Ans.     v  =  1203  ft.  per  sec. 

5.  A  soft  wood  block  weighing  5  pounds 
is    suspended    by   vertical   cords  5  feet  in 

length.     A   bullet  weighing  0.02  pound  is  jpIG   318 

fired   horizontally   into   the  block  and  re- 
mains imbedded  in  it.     The  block  swings  through  a  vertical  angle  of  35 
Find  the  velocity  of  the  bullet. 


II 


233.  Collision  of  Elastic  Bodies. — When  two  elastic  bodies 
collide,  both  bodies  suffer  a  change  of  form.  Their  centers  of 
mass  continue  to  approach  each  other 
for  a  little  time  and  both  bodies  are 
compressed  at  the  surface  of  contact. 
If  one  of  the  bodies  is  fixed  so  as  to  be 
practically  stationary,  the  other  body 
will  come  to  rest  and  then  rebound. 
If  both  bodies  are  free  to  move,  there 
comes  a  time  when  each  is  stationary 
relative  to  the  other,  and  both  are 
moving  with  the  same  velocity  relative 
to  the  earth.  Fig.  319  represents  three 
stages  of  the  collision  of  two  bodies. 
In  Fig.  319,  I,  the  bodies  are  moving 
toward  each  other  at  the  beginning  of 
contact.  In  Fig.  319,  II,  the  centers 
Oi  and  0%  are  nearest  to  each  other. 
The  bodies  are  stationary  relative  to 
each  other  and  are  moving  with  a  velocity  v  relative  to  the 
earth.  It  is  assumed  that  the  momentum  of  m\  is  the  greater; 
the  common  velocity  is,  therefore,  toward  the  right.  The  interval 
from  the  beginning  of  collision  to  the  condition  of  Fig.  319, 


FIG.  319. 


384 


MECHANICS 


[ART.   234 


II,  is  called  the  compression  stage  of  the  collision.  The  common 
velocity  v  at  the  end  of  the  compression  stage  is  found  in  the 
same  way  as  if  the  bodies  were  inelastic. 

After  the  bodies  have  reached  the  end  of  the  compression 
stage,  their  elasticity  causes  them  to  separate.  The  interval 
from  the  end  of  the  compression  stage  to  the  time  when  the 
bodies  cease  to  be  in  contact  with  each  other  is  called  the  restitu- 
tion stage.  Fig.  319,  III,  shows  the  bodies  at  the  end  of  the 
restitution  stage. 

If  the  bodies  are  perfectly  elastic,  the  impulse  during  resti- 
tution is  equal  to  the  impulse  during  compression  and  the  change 
of  momentum  of  each  body  during  restitution  is  equal  to  the 
change  during  compression.  With  imperfectly  elastic  bodies 
the  change  of  momentum  during  restitution  is  less  than  the 
change  during  compression.  The  ratio  of  the  impulse  during 
restitution  to  the  impulse  during  compression  is  called  the 
coefficient  of  restitution.  When  the  bodies  are  perfectly  elastic, 
the  coefficient  of  restitution  is  unity.  The  coefficient  of  restitu- 
tion is  expressed  by  the  letter  e. 

Problems 

1.  A  body  is  thrown  against  a  fixed 
wall  with  a  speed  of  60  feet  per  second 
and  rebounds  with  a  speed  of  45  feet 
per  second.     Find  the  coefficient  of 
restitution.  Ans.     e  =  0.75. 

2.  A  ball  is   dropped   4   feet  upon 
a  horizontal  surface  and  rebounds  3 
feet.  Find  the  coefficient  of  restitution. 

Ans.     e  =  0.866. 

3.  A  body  is  dropped  10  feet  to  a 
horizontal  surface.     If  the  coefficient 
of  restitution  is  0.8,  how  high  will  it 
rebound? 


II 


III 


The  behavior  of  a  pair  of 
""  elastic  bodies  during  collision  is 
best  studied  by  a  numerical 
example  rather  than  by  the  use 
of  literal  formulas.  Fig.  320,  I, 
represents  a  mass  of  25  pounds 
moving  east  with  a  velocity  of  40 
feet  per  second,  which  is  overtaking  a  mass  of  15  pounds  moving 
east  with  a  velocity  of  20  feet  per  second.  Fig.  320,  II,  shows 
the  two  bodies  at  the  end  of  the  compression  stage.  It  may 


FIG.  320. 


CHAP.  XXV]       MOMENTUM  AND  IMPULSE  385 

be  assumed  that  the  bodies  are  photographed  and  that  the 
plate  is  shifted  vertically  between  exposures.  The  initial 
momentum  is 

25  X  40  +  15  X  20  =  1300. 
v  =  32.5  feet  per  second. 

As  a  condition  of  the  problem,  it  will  be  assumed  that  the 
coefficient  of  restitution  is  0.8.  The  change  of  velocity  of  the 
25-pound  mass  during  compression  is  32.5  —  40  =  —7.5  feet 
per  second.  The  change  of  velocity  of  this  mass  during  restitu- 
tion is  0.8  (  —  7.5)  =  —  6  feet  per  second.  The  final  velocity 
of  the  25-pound  mass  is 

v  =  32.5  —  6  =  26.5  feet  per  second  east. 

The  change  of  velocity  of  the  15-pound  mass  during  compression 
is  12-.5  feet  per  second.  Its  change  during  restitution  is  10  feet 
per  second.  The  final  velocity  of  the  15-pound  mass  is 

v  =  32.5  H-  10  =  42.5  feet  per  second  east. 
The  results  may  be  checked  by  means  of  the  final  momentum; 

25  X  26.5  =    662.5 

15  X  42.5  =    637.5 


Total  momentum  =  1300. 

Problems 

1.  A  mass  of  40  pounds  moving  east  with  a  velocity  of  30  feet  per  second 
meets  a  mass  of  20  pounds  moving  west  with  a  velocity  of  45  feet  per  second. 
The  coefficient  of  restitution  is  0.6.     What  is  the  velocity  of  each  body  after 
collision?  Ans.     10  ft.  per  sec.  west;  35  ft.  per  sec.  east. 

2.  A  mass  of  20  pounds  moving  east  40  feet  per  second  meets  a  mass  of 
5  pounds  moving  west  50  feet  per  second.     After  collision,  the  20-pound  mass 
moves  east  8.5  feet  per  second.     By  equating  the  momenta,  find  the  velocity 
of  the  5-pound  mass  after  collision.         Ans.     v  =  76  ft.  per  sec.  east. 

3.  In  Problem  2,  what  is  the  common  velocity  at  the  end  of  the  compres- 
sion stage  and  what  is  the  coefficient  of  restitution? 

Ans.     v  =  22  ft.  per  sec.  east;  e  =  0.75. 

4.  A  mass  of  15  pounds  moving  east  60  feet  per  second  overtakes  a 
mass  of  10  pounds  moving  east  20  feet  per  second.     After  collision,  the 
mass  of  10  pounds  moves  east  65.6  feet  per  second.     Find  the  final  velocity 
of  the  mass  of  15  pounds  and  find  the  coefficient  of  restitution. 

6.  A  baseball  moving  100  feet  per  second  is  struck  by  a  bat  moving  60 
feet  per  second  in  the  opposite  direction.  If  the  bat  weighs  6  times  as  much 
as  the  ball,  and  the  coefficient  of  restitution  is  0.9,  with  what  velocity  will 
the  ball  leave  the  bat?  Ans.  133.6  ft.  per  sec. 

25 


386  MECHANICS  [ART.  234 

6.  In  Problem  1,  what  is  the  energy  of  each  mass  before  and  after  collision; 
what  is  the  total  loss  of  energy;  and  what  is  the  ratio  of  the  loss  to  the  original 
energy?  Ans.     Loss  =  621.6ft.  Ib.  =  62.7%  of  initial  energy. 

7.  Solve  Problem  1  if  the  bodies  are  perfectly  elastic,  and  find  the  energy 
loss.        Ans.  20  ft.  per  sec.  west;  55  ft.  per  sec.  east.     No  energy  lost. 

234.  Moment  of  Momentum. — If  a  body  of  mass  m  is  moving 

with  linear  velocity  v,  its  momentum  is  mv.     If  0,  Fig.  321, 

is  an  axis  perpendicular  to  the  plane  of  the  paper,  and  if  r  is 

the   length   of  the   line    perpendicular   to 

o  *V     this  axis  and  to  the  line  of  motion  of  the 

X  T^     mass  m,  the  product  of   the   momentum 

^/\  /  of  m  multiplied  by   the   length   r   is   the 

x:y  'moment  of   momentum   of  m  with  respect 

to   the   axis.     When    a   body   is   rotating 

about  an  axis,   the  linear  velocity  of  an 

element  dm  at  a  distance  r  from  the  axis  is  given  by  the- 
equation 

v  =  rco  (1) 

The  momentum  of  this  element  is  rudm  and  the  moment  of 
momentum  with  respect  to  the  axis  of  rotation  is  r2udm.  The 
total  moment  of  momentum  is 

a)frzdm  =  co/.  (2) 

The  moment  of  momentum  with  respect  to  an  axis  of  a  body 
which  is  rotating  about  that  axis  is  the  product  of  the  moment 
of  inertia  of  the  body  multiplied  by  its  angular  velocity.  A 
comparison  of  moment  of  momentum  with  momentum  shows 
that  moment  of  inertia  replaces  mass,  and  angular  velocity 
replaces  linear  velocity. 

To  change  the  momentum  of  a  body,  force  must  be  applied 
from  an  outside  body.  Likewise,  to  change  the  moment  of 
momentum  of  a  body,  torque  must  be  applied  from  some  out- 
side body  or  system  of  bodies.  From  Formula  XXX 

la  _  I  dwt  (  . 

7  ~g  Tt' 

;  (4) 

•       Tt-+C,  §  (5) 

in  which  C  is  an  integration  constant.  The  product  Tt  may  be 
called  the  impulse  of  the  torque. 


CHAP.  XXV]       MOMENTUM  AND  IMPULSE 


387 


The  moment  of  momentum  of  a  rotating  body  may  be  changed 
by  changing  the  angular  velocity  or  by  changing  the  configuration 
of  the  body  in  such  a  way  as  to  alter  its  moment  of  inertia.  Fig. 
322  shows  a  frame  arranged  to  rotate  about  a  vertical  axis.  The 
frame  carries  two  equal  bodies  A  and  B.  These  bodies  are 
connected  to  cords  which  run  over  pulleys  and  are  attached  to  the 
sleeve  F  on  the  vertical  shaft.  Suppose  that  the  frame  is  rotating 
with  an  angular  velocity  co  when  A  and  B  are  in  the  positions 
shown  in  the  figure.  If  the  sleeve  F  is  now  pulled  down  by  some 
outside  force,  or  by  its  own  weight,  and  draws  A  and  B  in  toward 
the  axis,  the  moment  of  inertia  of  the  system  is  reduced.  Since 


FIG.  322. 

the  moment  of  momentum  remains  constant,  the  angular  velocity 
of  the  frame  must  increase.  If  the  sleeve  is  lifted  and  the 
bodies  A  and  B  are  allowed  to  move  out  to  the  original  positions, 
the  moment  of  inertia  is  again  increased  and  the  angular  velocity 
is  correspondingly  diminished. 

In  Fig.  323,  a  mass  m  is  supposed  to  be  revolving  in  a  horizontal 
plane  about  a  vertical  axis  perpendicular  to  the  plane  of  the 
paper  at  0.  The  mass  is  at  a  distance  r  from  the  axis  and  may 
be  regarded  as  supported  by  a  frictionless  horizontal  plane  and 
held  by  a  cord  attached  to  the  axis.  If  the  cord  is  released  at 
the  axis,  the  mass  will  move  along  the  line  of  the  tangent  to  its 
path  at  the  moment  of  release.  If  the  angular  velocity  is  coi,  the 
linear  velocity  in  the  tangent  line  will  be  rcoi.  Suppose  that  the 
cord  is  stopped  at  the  axis  when  the  mass  is  in  the  position  of 
Fig.  323,  II,  at  a  distance  R  from  the  axis.  The  linear  velocity  of 


388 


MECHANICS 


[ART.   234 


the  mass  is  still  rcoi.  The  component  of  this  linear  velocity  per- 
pendicular to  the  radius  R  is  ran  cos  6.  Its  angular  velocity 
about  the  axis  is  this  component  of  the  linear  velocity  divided 
by  R 

7*00 1  COS  0 


co  = 


R 


(6) 


Since  cos  6  =  H> 


r2coi, 


(7) 
(8) 

in  which  co  is  the  angular  velocity  at  the  new  position.     The 
moment  of  inertia  at  the  first  position  is  rar2  and  at  the  second 


FIG.  323. 

position  is  mR2.  Multiplying  both  sides  of  Equation  (8)  by 
m, 

/CO    =    71W1,  (9) 

in  which  /i  is  the  moment  of  inertia  at  the  first  position  at  a 
distance  r  from  the  axis,  and  /  is  the  moment  of  inertia  at  the 
second  position  at  a  distance  R  from  the  axis.  It  may  be  assumed 
that  the  dimensions  are  relatively  so  small  that  the  moment  of 
inertia  of  the  body  with  respect  to  an  axis  through  the  center  of 
mass  is  negligible.  This  assumption,  however,  is  not  necessary, 
since  the  body  may  be  considered  as  composed  of  a  great  number 
of  small  elements,  each  of  which  has  its  own  r  and  R. 

Equation  (9)  gives  an  independent  illustration  of  the  fact 
that  the  moment  of  momentum  is  not  altered  when  the  configura- 
tion is  changed  by  forces  which  exert  no  torque  about  the  axis. 
On  the  other  hand,  the  kinetic  energy  of  the  system  is  changed. 


CHAP.  XXV]       MOMENTUM  AND  IMPULSE  389 

The  kinetic  energy  varies  as  the  square  of  the  angular  velocity 
while  the  moment  of  momentum  varies  as  the  first  power  of  the 
angular  velocity.  When  the  moment  of  inertia  is  doubled,  the 
angular  velocity  becomes  one-half  as  great  and  the  kinetic  energy 
becomes  one-half  as  great.  In  the  system  of  Fig.  323,  the  kinetic 
energy  is  expended  when  the  radial  motion  is  stopped  by  the 
inelastic  cord.  If  the  body  is  brought  back  toward  the  axis, 
considerable  force  must  be  exerted  on  the  cord.  The  work  done 
on  the  system  through  the  cord  represents  the  increase  of  kinetic 
energy. 

Example 

A  frame  similar  to  Fig.  322  carries  two  masses,  each  of  which  weighs 
3  pounds.  The  moment  of  inertia  of  the  frame  about  its  axis,  together  with 
the  moment  of  inertia  of  each  mass  about  an  axis  through  its  center  of  mass 
parallel  to  the  axis  of  the  frame  is  2  units  (expressed  in  pounds  and  feet). 
When  the  masses  are  each  1  foot  from  the  vertical  axis,  the  system  is  rotating 
with  an  angular  velocity  of  10  radians  per  second.  The  sleeve  attached  to 
the  cords  is  then  raised  and  the  masses  move  to  positions  2  feet  from  the  axis. 
Find  the  angular  velocity  and  the  change  of  kinetic  energy. 

In  the  first  position 

/i  =  2  +  6  =  8; 

Moment  of  momentum  =  8  X  10  =  80. 
In  the  second  position 

/2  =  2  +  6  X  4  =  26; 
26o>2  =  80; 

co2  =  3.08  radians  per  sec. 

In  the  first  position,  the  kinetic  energy  is  12.43  foot-pounds.  In  the  second 
position,  the  kinetic  energy  is  4.86  foot  pounds. 

Problems 

1.  In  the  above  example  the  two  masses  are  drawn  in  till  their  centers 
of  gravity  are  each  6  inches  from  the  axis.     Find  the  angular  velocity. 

Ans.     w  =  22.86  radians  per  second. 

2.  A  circular  disk  2  feet  in  diameter  and  weighing  4  pounds  rotates  about 
a  vertical  axis  with  an  angular  velocity  of  6  radians  per  second.     A  second 
disk  1  foot  in  diameter  and  weighing  16  pounds  is  coaxial  with  the  first  disk. 
This  second  disk  is  dropped  a  short  distance  upon  the  first  disk  and  adheres 
to  it.     If  the  second  disk  is  initially  stationary,  what  is  the  common  velocity 
of  the  two  when  they  rotate  together?     Ans.     co  =  3  radians  per  second. 

3.  Solve  Problem  2  if  the  axis  of  the  second  disk  is  6  inches  from  the  axis 
of  rotation. 

4.  Solve  Problem  2  if  the  second  disk  has  an  initial  velocity  of  10  radians 
per  second  in  the  same  direction  as  the  first  disk. 


390 


MECHANICS 


[ART.   234 


FIG.  324. 


5.  Figure  324  represents  a  frame  arranged  to  rotate  about  a  vertical  axis 
with  little  friction.  A  mass  m  rolls  on  a  curved  arm.  The  moment  of 
inertia  of  the  frame  about  its  axis  together  with  the  moment  of  inertia  of  m 

about  a  vertical  axis  through  its  center 
of  gravity  is  /.  When  the  mass  m  is  at 
a  distance  r\  from  the  axis  of  rotation, 
the  angular  velocity  of  the  frame  is  coi 
radians  per  second.  Find  the  form  of  the 
arm  in  order  that  the  mass  m  may  be  in 
equilibrium  in  any  position. 

.    mru2    . 
I  he  centrifugal  iorce  is >  in 

a 

which  w  is  the  angular  velocity 
when  the  center  of  mass  is  at  a 
distance  r  from  the  axis. 

When  the  center  of  mass  is  at  a 
distance    r\    from    the    axis,    the 

moment  of  inertia  is  7  +  mr\ ;  and  when  m  is  at  a  distance  r  from 
the  axis,  the  moment  of  inertia  is  I  +  mr2-  If  w  is  the  angular 
velocity  when  the  center  of  mass  of  m  is  at  a  distance  r  from  the 
axis,  the  equation  of  moment  of  momentum  is 

(7  +  mrj)coi  =  (/  +  mr2)co. 
2        (7+  mrg)X 
(7  +  rar2)2 

By  the  method  of  virtual  work,  the  equation  of  the  surface  of 
equilibrium  is  given  by  the  expression 

Hdx  +  Vdy  =  0. 

The  centrifugal  force  is  H  and  the  weight  of  m  is  V.  Since  the 
weight  is  downward,  V  =  —  m.  Substituting  the  value  of 
co2  in  the  expression  for  the  centrifugal  force  and  replacing  dx 
by  dr,  the  equation  of  the  surface  becomes, 


-  (1  + 


If  y  =  0  when  r 


2gm(I 


_  (I  +  mrfX  . 

2gm 

(7+mr2)o)2/        7  +  mrf 
2gm         \      ~  I  +  — 2 


mr 


CHAP.  XXV]       MOMENTUM  AND  IMPULSE 


391 


When  the  angular  velocity  exceeds  on  by  ever  so  little,  the 
mass  m  moves  out  to  the  extremity  of  the  arm  and  the  angular 
velocity  falls.  The  kinetic  energy  of  rotation  is  diminished, 
while  the  potential  energy  is  increased.  When  the  velocity 
is  slightly  diminished,  the  mass  m  moves  inward  toward  the 
axis,  and  the  angular  velocity  increases. 

235.  Center  of  Percussion. — The  center  of  percussion  of  a 
body  with  respect  to  an  axis  is  the  point  at  which  a  sudden 
impulse  normal  to  the  plane  through  the  axis  and  the  center  of 
mass  may  be  given  to  the  body 
without  causing  an  additional  re- 
action at  the  axis  in  the  direction 
of  the  impulse.  Fig.  325  represents 
a  bar  hinged  at  the  top.  If  a  blow 
perpendicular  to  its  length  is  de- 
livered to  the  bar  at  the  middle, 
the  bar  tends  to  move  toward  the 
left  parallel  to  itself.  In  order 
that  it  may  rotate  around  the 
hinge,  the  hinge  must  exert  a  force 
toward  the  right.  If  the  blow  is 
delivered  near  the  bottom,  as  shown 
in  Fig.  325,  II,  the  bar  tends  to 
rotate  in  a  clockwise  direction  about 
some  axis  below  the  hinge.  The 
hinge  must  then  exert  a  reaction 
toward  the  left  in  the  direction  of 
the  impulse.  Between  the  middle 
of  the  bar  and  the  bottom  there 

is  some  point  at  which  a  blow  may  be  struck  without  developing 
any  horizontal  reaction  in  the  hinge.  This  point  is  the  center 
of  percussion.  Students  who  play  baseball  are  familiar  with 
the  effects  of  failure  to  strike  a  ball  at  the  center  of  percussion 
of  the  bat. 

If  P  is  the  force  on  a  body  such  as  the  bar  of  Fig.  325,  a  is  the 
linear  acceleration  of  the  center  of  mass,  and  N  is  the  normal 
component  of  the  hinge  reaction  opposite  the  direction  of  P. 
Formula  XXII  gives  the  equation 


FIG.  325. 


P  -N  = 


ma 

T' 


(i) 


392  MECHANICS  [ART.   23.5 

If  N  =  0, 

P  =  ^  (2) 

If  the  force  P  acts  at  a  distance  /  from  the  axis  of  rotation,  its 
torque  is  PL  By  Formula  XXX 

Pl=j-  •  (3) 

When  the  angular  acceleration  is  a,  the  linear  acceleration  of 
the  center  of  mass  is 

a  =  ra.  (4) 

Substituting  P  =  -    -  in  Equation     (3) 

m  al      la  .  N 

(5) 

Q          9 
mrl  =  I  =  m/b2;  (6) 

JU2 

*  =  r  (7) 

Equation  (7)  states  that  the  center  of  percussion  is  located  at 
a  distance  from  the  axis  equal  to  the  square  of  the  radius  of 
gyration  divided  by  the  distance  from  the  axis  to  the  center  of 
gravity.  A  comparison  with  Art.  226  shows  that  the  center 
of  percussion  lies  on  the  axis  of  oscillation. 

Problems 

1.  A  uniform  rod  of  length  L  is  suspended  on  an  axis  through  one  end 
perpendicular  to  its  length.     Find  the  center  of  percussion. 

Ans.     f  L  from  the  axis  of  suspension. 

2.  Find  the  center  of  percussion  of  a  uniform  rod  suspended  by  an  axis 
at  one-third  its  length  from  one  end. 

3.  Find  the  center  of  percussion  of  a  uniform  rod  for  an  axis  at  three- 
eighths  the  length  from  one  end. 

Ans.     £  L  from  the  other  end. 

4.  A  bat  in  the  form  of  a  frustrum  of  a  cone  is  1.5  inches  in  diameter  at 
one  end,  3  inches  in  diameter  at  the  other  end,  and  3  feet  long.     It  is  gripped 
4  inches  from  the  small  end.     If  the  bat  is  swung  around  the  point  at  which 
it  is  gripped,  where  should  it  strike  the  ball  in  order  that  the  jar  on  the  hand 
may  be  a  minimum? 

6.  A  triangular  paddle  is  made  of  a  board  of  uniform  thickness.  Where 
is  the  center  of  percussion  when  the  axis  of  suspension  passes  through  the 
vertex  parallel  to  the  face  of  the  board,  and  where  is  the  center  of  percussion 
when  the  axis  of  suspension  passes  through  the  vertex  perpendicular  to  the 
face  of  the  hoard? 


CHAP.  XXV]       MOMENTUM  AND  IMPULSE  393 

The  center  of  percussion  may  also  be  defined  as  the  point 
about  which  the  moment  of  momentum  is  zero.  In  Fig.  326, 
0  is  the  axis  of  rotation  perpendicular  to  the  plane  of  the  paper 
and  C  is  a  parallel  axis  through  the  center  of  percussion.  The 
center  of  gravity  of  the  body  lies  in  the 
plane  through  the  axis  of  rotation  and  C. 
An  element  of  mass  dm  at  a  distance  r 
from  the  axis  of  rotation  has  a  linear 
velocity  of  ro>.  The  moment  of  momentum 
of  this  element  with  respect  to  the  axis 
through  C  is  the  linear  momentum  mul- 
tiplied by  the  length  CB  drawn  from  C 
perpendicular  to  the  direction  of  the  linear 
velocity.  If  I  is  the  distance  from  0  to  C 
and  0  is  the  angle  between  r  and  OC, 

CB  =  (I  -  r  sec  6)  cos  0  =  I  cos  6  -  r.  (8) 

The  moment  of  momentum  of  the  element  dm  is 

moment  of  momentum  =  co(Z  r  cos  6  —  r2)dm.  (9) 

Integrating  Equation  (9)  and  equating  to  zero, 

Ifr  cos  B  dm  =  frzdm.  (10) 

Since  fr  cos  6  dm  is  the  moment  of  the  entire  mass  with  respect 
to  the  plane  through  the  axis  of  suspension  perpendicular  to 
OC, 

fr  cos  6  dm  =  rm,  (11) 

in  which  r  is  the  distance  of  the  center  of  mass  from  the  axis 
of  suspension.  Equation  (10)  then  becomes 

lrm=kzm',  (12) 


236.  Summary.  —  The  product  of  the  mass  of  a  body  multi- 
plied by  its  linear  velocity  is  called  its  momentum.  The  momen- 
tum of  a  body  or  system  of  bodies  is  constant  unless  changed 
by  the  application  of  force  from  another  body  or  system. 
Momentum  is  a  vector  quantity. 

The  product  of  force  multiplied  by  time  is  called  the  impulse 
of  the  force.  The  impulse  on  a  body  is  proportional  to  the 


394  MECHANICS  [ART.  326 

change  of  momentum.  With  absolute  units,  the  impulse  is 
numerically  equal  to  the  change  of  momentum.  With  gravita- 
tional units,  the  impulse  is  numerically  equal  to  the  change  of 
momentum  divided  by  g. 

Since  action  and  reaction  are  equal,  when  two  bodies  exert 
force  on  each  other,  their  impulses  are  equal.  As  a  result  the 
bodies  suffer  equal  and  opposite  changes  of  momentum.  The 
total  change  of  momentum  of  two  bodies  due  to  forces  which  act 
between  the  bodies  is  zero. 

When  inelastic  bodies  collide,  their  velocity  after  impact 
is  calculated  from  the  fact  that  the  total  momentum  is  not 
changed.  When  elastic  bodies  collide,  they  approach  for  a 
short  time.  At  the  end  of  this  interval,  which  is  called  the 
stage  of  compression,  the  bodies  are  moving  with  equal  velocities 
in  the  same  direction.  The  common  velocity  at  the  end  of  the 
compression  stage  is  the  same  as  it  would  be  if  the  bodies  were 
inelastic.  After  they  reach  the  end  of  the  compression  stage, 
the  elasticity  of  the  bodies  causes  them  to  separate.  The  interval 
from  the  end  of  the  compression  stage  until  the  bodies  cease  to 
touch  is  the  stage  of  restitution.  The  ratio  of  the  impulse 
during  restitution  to  the  impulse  during  compression  is  the 
coefficient  of  restitution.  When  the  coefficient  of  restitution  is 
unity,  the  bodies  are  perfectly  elastic  and  no  energy  is  lost  in 
the  collision. 

The  product  of  the  momentum  of  a  body  multiplied  by  the 
perpendicular  distance  of  its  line  of  action  from  a  given  axis 
is  the  moment  of  momentum  of  the  body  with  respect  to  that 
axis.  The  moment  of  momentum  can  be  changed  only  by  torque 
from  an  outside  body. 

The  center  of  percussion  with  respect  to  an  axis  is  the  point 
at  which  an  impulse  may  be  given  to  the  body  in  a  direction 
normal  to  the  plane  through  the  axis  and  the  center  of  gravity 
without  causing  any  change  in  the  reaction  of  the  axis  in  the 
direction  of  the  impulse.  The  center  of  percussion  falls  on  the 
axis  of  oscillation.  The  moment  of  momentum  of  a  rotating 
body  about  an  axis  through  the  center  of  percussion  parallel 
to  the  axis  of  rotation  is  zero. 


CHAPTER  XXVI 
ENERGY  TRANSFER 

237.  Units  of  Energy. — In  English-speaking  countries,  energy 
is  measured  by  engineers  in  foot-pounds.  Where  the  metric 
system  is  used,  the  engineering  unit  of  energy  is  the  kilogram- 
meter.  Since  1  kilogram  is  equal  to  2.20462  pounds,  and  1 
meter  is  equal  to  3.28083  feet, 

1  kilogram-meter  =  7.2330  foot-pounds; 

1  foot-pound          =  0.1 3826  kilogram-meters. 

For  some  purposes,  energy  is  expressed  in  gram-centimeters. 
1  foot-pound  =  13,826  gram-centimeters. 

For  scientific  purposes,  work  and  energy  are  measured  in 
ergs.  An  erg  is  the  work  done  by  the  force  of  one  dyne  when 
the  point  of  application  moves  one  centimeter  in  the  direction 
of  the  force.  At  the  standard  location,  the  acceleration  of  grav- 
ity is  32.174  X  30.4801  =  980.67  cm.  per  sec.  per  sec. 

1  gram-centimeter  =  980.67  ergs. 

1  foot-pound  =  13,558,700  ergs. 
Physicists  use  the  joule  as  a  unit  of  work  or  energy. 

1  joule  =  10,000,000  ergs.  =  107  ergs. 
1   foot-pound  =  1.35587   joules. 

Problems 

1.  A  40-pound   mass  is  pulled  up  a  30-degree  inclined  plane  by  a  force 
parallel  to  the  plane.     The  coefficient  of  friction  is  0.1.     How  much  work  is 
done  and  how  much  is  the  potential  energy  increased  when  the  body  is 
moved  50  feet?     Ana.     1173.2  ft.-lb.;  1000  ft.-lb. 

2.  Express  the  answers  to  Problem  1  in  kilogram-meters. 

3.  A  4-pound  mass  is  lifted  vertically  a  distance  of  25  feet  at  a  place  where 
g  =  32.2.     Find  the  work  in  joules.     Ans.     135.69  joules. 

395 


396  MECHANICS  [ART.   238 

238.  Power. — Rate  of  working  is  called  power.     The  unit  of 
power  in  English-speaking  countries  is  the  horsepower. 

1  horsepower  =  550  foot-pounds  per  second. 
1  horsepower  =  33,000  foot-pounds  per  minute. 

For  electrical  measurement,  the  watt  is  the  unit  of  power. 

1  watt  =  1  joule  per  second  =  107  ergs  per  second. 
1  horsepower  =  550  X  1.35587  =  745.7  watts. 

It  is  customary  to  use 

1  horsepower  =  746  watts. 

For  direct  currents  and  for  alternating  currents  in  non-induc- 
tive circuits,  the  power  in  watts  is  equal  to  the  product  of  the 
potential  difference  of  the  terminals  of  the  circuit  in  volts  multi- 
plied by  the  current  in  amperes. 

Watts  =  El  =  volts  X  amperes. 
Since  E  =  RI  in  a  non-inductive  circuit, 

watts  =  PR, 
in  which  R  is  the  resistance  in  ohms. 

Problems 

1.  A  tank,  20  feet  square  and  30  feet  deep,  is  filled  with  water  from  a 
source  which  is  40  feet  below  the  bottom  of  the  tank.     If  the  intake  pipe 
enters  the  bottom  of  the  tank,  what  is  the  horsepower  required  to  fill  it  in 
50  minutes.     Ans.     25  hp. 

2.  The  pump  in  Problem  1  has  an  efficiency  of  80  per  cent  and  is  driven 
by  a  motor  which  has  an  efficiency  of  90  per  cent.     How  many  kilowatts  are 
required  to  drive  the  motor?     Ans.     25.9  kilowatts. 

3.  A  conical  tank  is  20  feet  in  diameter  at  the  top  and  30  feet  deep.     It  is 
filled  with  water  from  a  source  40  feet  below  the  bottom  by  means  of  a  pump 
of  75  per  cent  efficiency  which  is  driven  by  a  220-volt  motor  of  90  per  cent 
efficiency.     If  the  tank  is  filled  in  30  minutes  through  a  pipe  which  enters  at 
the  bottom,  what  is  the  average  current  supplied  to  the  motor? 

4.  How  many  horsepower  equal  one  kilowatt? 

6.  A  flywheel  6  feet  in  diameter  is  making  300  revolutions  per  minute. 
The  tension  on  one  side  of  the  belt  is  1800  pounds  and  the  tension  on  the 
other  side  is  300  pounds.  Find  the  horsepower  transmitted.  Ans.  25.7  hp. 

6.  If  T  is  the  torque  in  foot-pounds  in  a  shaft,  show  that  the  work  per 
revolution  is  2wT, 

239.  Mechanical  Equivalent  of  Heat. — Energy  can  neither  be 
created  nor  destroyed.     When  it  apparently  disappears,  it  merely 
changes  its  form.    The  statement  that  the  total  energy  of  a 


CHAP.  XXVI]  ENERGY  TRANSFER  397 

closed  system  remains  unchanged  is  called  the  Law  of  Conserva- 
tion of  Energy. 

The  most  common  change  of  energy  is  the  change  to  the  form  of 
heat  energy.  When  friction  takes  place  between  two  bodies, 
heat  is  generated.  The  temperature  of  the  bodies  may  be  so 
raised  by  the  heat  of  friction  that  the  bodies  are  melted  or  ignited. 

Heat  energy  is  measured  in  British  thermal  units  or  in  calories. 
A  British  thermal  unit  is  the  heat  required  to  raise  the  tempera- 
ture of  one  pound  of  water  one  degree  Fahrenheit.  Since  the 
specific  heat  of  water  varies  slightly  with  the  temperature,  a 
British  thermal  unit  is  defined  more  accurately  as  the  amount  of 
heat  required  to  raise  the  temperature  of  one  pound  of  water  from 
32  degrees  to  33  degrees  Fahrenheit. 

A  calorie  is  the  amount  of  heat  required  to  raise  the  tem- 
perature of  one  kilogram  of  water  one  degree  Centigrade.  A 
calorie  is  defined  accurately  as  the  amount  of  heat  required  to 
raise  the  temperature  of  one  kilogram  of  water  from  0  degrees  to 
1  degree  Centigrade. 

A  small  calorie  is  the  amount  of  heat  required  to  raise  the 
temperature  of  1  gram  of  water  from  0  degrees  to  1  degree  Centi- 
grade. The  small  calorie  is  called  a  calorie  by  physicists. 

The  experiments  of  Joule  and  Rowland  have  shown  that 
427  kilogram-meters  of  work  will  raise  -the  temperature  of  1 
kilogram  of  water  1  degree  Centigrade.  This  figure  is  called 
the  mechanical  equivalent  of  heat. 

1   large  calorie   =  427  kilogram  meters  =  J. 
1  British  thermal  unit  =  778  foot-pounds. 

Problems 

1.  If  the  work  done  in  lifting  1  kilogram  a  distance  of  427  meters  is 
sufficient  to  raise  the  temperature  of  1  kilogram  of  water  1  degree  Centigrade, 
how  high  will  a  mass  of  1  pound  be  lifted  by  the  energy  which  is  required  to 
raise  the  temperature  of  1  pound  of  water  1  degree  Centigrade? 

Ans.     h  =  1400    ft. 

2.  If  the  energy  required  to  raise  the  temperature  of  1  pound  of  water 
1  degree  Centigrade  is  sufficient  to  lift  1  pound  a  vertical  distance  of  1400 
feet,  how  high  will  the  energy  required  to  raise  the  temperature  of  1  pound  of 
water  1  degree  Fahrenheit  lift  1  pound? 

3.  A  friction  brake  which  is  absorbing  30  horsepower  is  cooled  by  water 
at  20  degrees  Centigrade.     The  water  is  evaporated  from  and  at  100  degrees 
Centigrade.     How  many  pounds  of  water  are  required  per  hour? 

Ans.    68.7  Ib, 


398 


MECHANICS 


[ART.   239 


4.  A  sled  weighing  1200  pounds  is  drawn  along  a  horizontal  ice  surface  by 
a  horizontal  pull  of  70  pounds.  If  the  temperature  of  the  ice  is  32  degrees 
Fahrenheit,  how  much  ice  is  melted  when  the  sled  moves  100  feet? 

6.  An  iron  shot  moving  at  a  speed  of  1200  feet  per  second  is  stopped  by 
striking  an  object.  If  the  specific  heat  of  iron  is  0.11  and  if  one-half  of  the 
energy  goes  to  heat  the  shot,  how  high  is  its  temperature  raised? 

6.  A  meteorite  strikes  the  air  with  a  velocity  of  6  miles  per  second  rela- 
tive to  the  earth.     If  the  specific  heat  is  0.11  and  if  one-half  of  the  energy 
goes  to  heat  the  meteorite,  how  high  is  its  temperature  raised? 

7.  The  drum  of  Fig.  327  is  mounted  on  a  smooth  pivot.     The  vertical 
shaft  at  the  top  drives  a  set  of  paddles  inside  the  drum.     The  drum  is  rilled 
with  water  which  is  heated  by  the  mechanical  energy  of  the  rotating  paddles. 


FIG.  327. 

The  torque  is  measured  by  means  of  two  cords  which  are  wound  round 
the  drum  and  run  over  smooth  pulleys.  The  water  equivalent  of  heat  of  the 
apparatus  was  found  to  be  2.42  pounds.  With  17.58  pounds  of  water  in  the 
drum,  the  temperature  was  raised  from  50°  Fahr.  to  74°  Fahr.  while  the  shaft 
made  4000  revolutions.  The  average  torque  during  the  experiment  was 
10.8  foot-pounds.  Find  the  mechanical  equivalent  of  heat. 

8.  How   many  kilowatts  are  required  to  raise  the  temperature  of  20 
pounds  of  water  from  62  degrees  to  boiling  in  30  minutes? 

9.  How  many  kilowatt-hours  are  required  to  raise  10  kilograms  of  water 
from  20  degrees  C.  to  100  degrees  C.? 

10.  Coal  with  a  calorific  value  of  13000  B.t.u  per  pound  is  used  with  a 
boiler  of  70  per  cent  efficiency  and  an  engine  of  15  per  cent  efficiency.     How 
many  pounds  of  coal  are  required  per  horsepower-hour?    Ans.     1.87  Ib. 


CHAP.  XXVI] 


ENERGY  TRANSFER 


399 


240.  Power  of  a  Jet  of  Water. — When  a  liquid  flows  from  an 
orifice,  the  velocity  is  the  same  as  that  which  a  body  would 
acquire  in  falling  freely  from  a  height  equal  to  the  vertical  dis- 
tance of  of  the  orifice  below  the  surface  the  liquid.  In  Fig.  328, 

v  =  V2jh>  (1) 

in  which  h  is  measured  from  the  center  of  the  orifice  and  v  is  the 
velocity  at  the  center.  If  h  is  measured  from  any  other  point 
in  the  orifice,  the  velocity  v  is  the  velocity  at  that  point.  For  an 


I  *••  Contracted 

Vein 

FIG.  328. 

orifice  of  small  dimensions  compared  with  the  height  of  the 
water,  the  velocity  at  the  middle  agrees  with  the  average  velocity 
within  narrow  limits. 

The  velocity  of  Equation  (1)  is  called  the  theoretical  velocity. 
Fig.  328,  I,  shows  an  orifice  in  a  thin  plate  (sometimes  called  a 
frictionless  orifice).  Practically  no  energy  is  lost  in  an  orifice  of 
this  kind,  and  Equation  (1)  gives  the  true  velocity.  When  an 
orifice  of  this  kind  is  placed  in  a  plane  surface  of  dimensions 
considerably  greater  than  the  diameter  of  the  orifice,  the  momen- 
tum of  the  liquid  flowing  in  from  all  sides  tends  to  carry  it  in 
past  the  edge.  As  a  result  of  this  transverse  component,  the 
motion  at  the  surface  is  not  normal  to  the  plane  of  the  orifice  and 
the  jet  is  contracted  on  all  sides.  At  a  distance  from  the  orifice 
about  equal  to  its  radius,  all  the  liquid  is  moving  in  one  direction 
and  the  diameter  of  the  jet  is  considerably  smaller  than  that 


400  MECHANICS  [ART.  240 

of  the  orifice.  This  section  of  the  jet  is  called  the  contracted  vein. 
The  ratio  of  the  area  of  the  contracted  vein  to  the  area  of  the 
orifice  is  called  the  coefficient  of  contraction.  With  full  contrac- 
tion, the  coefficient  of  contraction  for  circular  orifices  is  about 
0.62. 

The  quantity  of  liquid  which  flows  from  an  orifice  in  one  unit 
of  time  is  equal  to  the  area  of  the  contracted  vein  multiplied 
by  the  average  velocity.  The  quantity  is  given  by  the  equation, 

Q  =  KAV2Jh,  (2) 

in  which  A  is  the  area  of  the  orifice,  and  K  is  a  coefficient  called 
the  coefficient  of  discharge.  The  coefficient  of  discharge  is  the 
ratio  of  the  quantity  actually  discharged  to  the  quantity  which 
would  flow  through  the  orifice  if  there  were  no  contraction  and 
the  actual  velocity  were  the  full  theoretical  velocity.  In  an 
orifice  in  a  thin  plate,  the  velocity  is  the  full  theoretical  velocity 
and  the  coefficient  of  discharge  is  equal  to  the  coefficient  of  con- 
traction. The  coefficient  of  contraction  for  an  orifice  of  this 
kind  is  0.62.  The  equation 

Q  =  Q.62A\/2gh 

gives  the  discharge  from  an  orifice  in  a  thin  plate. 

Figure  328,  II,  shows  a  gradually  converging  short  nozzle. 
Since  the  area  of  the  jet  is  equal  to  the  area  of  the  nozzle,  the 
coefficient  of  contraction  is  unity.  The  friction  slightly  reduces 
the  velocity  at  the  surface;  the  coefficient  of  discharge  is,  there- 
fore, a  little  less  than  unity.  With  the  proper  curvature,  K  is 
greater  than  0.99.  With  ordinary  nozzles,  K  is  about  0.95. 

The  power  of  a  jet  may  be  found  by  means  of  the  kinetic 
energy  per  pound  of  liquid.  Since  the  energy  is  equivalent  to 
the  work  done  in  lifting  a  pound  a  vertical  distance  of  h  feet,  the 
power  is  easiest  calculated  by  multiplying  the  number  of  pounds 
per  second  by  the  height  h. 

Example 

Water  flows  from  a  6-inch  orifice  in  a  thin  plate  under  a  head  of  60  feet. 
The  coefficient  of  contraction  is  0.62.  Find  the  horsepower  of  the  jet. 


v  =  V2  X  32.174  X  60  =  62.14  ft.  per  sec. 
Q  =  0.62  X  ~Q  X  62.14  =  7.582   cu.    ft.    per   sec. 

Since  one  cubic  foot  of  water  weighs  nearly  62.5  pounds,  the  energy  of 
each  cubic  foot  is  62.5  X  60  =  3750  ft.-lb.  The  total  energy  is  7.582  X 
3750  =  28,432  ft.-lb.  per  second.  This  is  51.7  horsepower. 


CHAP.   XXVI] 


ENERGY  TRANSFER 


401 


Pressure  of  a  liquid  or  gas  is  often  given  in  pounds  per  square  inch.     A 

£*  o    r 

column  of  water  1  inch  square  and  1  foot  high  weighs  -~r  =  0.434  pound. 

A  column  1  inch  square  and  2.30  feet  high  weighs  1  pound.  To  change  water 
pressure  in  pounds  per  square  inch  to  head  in  feet  of  water,  multiply  by 
2.30. 

Problems 

1.  Water  flows  from  a  4-inch  circular  orifice  in  a  thin  plate  under  a  pres- 
sure of  40  pounds  per  square  inch.     Find  the  horsepower  of  the  jet. 

Ans.     43.52  hp. 

2.  The  jet  from  the  orifice  of  Problem  1  drives  a  water  wheel  of  80  per 
cent  efficiency  which  drives  a  generator  of  90  per  cent  efficiency.     How 
many  kilowatts  are  generated?     Ans.     23.38  kw. 

3.  If  the  jet  from  a  given  orifice  delivers  60  hp.  when  the  head  is  40  feet 
of  water,  what  will  it  deliver  when  the  pressure  is  50  pounds  per  square  inch? 

241.  Work  of  an  Engine. — The  work  done  by  the  steam  in  an 
engine  during  one  stroke  is  the  product  of  the  total  pressure  on 
one  side  of  the  piston  multiplied  by  the  length  of  the  stroke. 

,-  Pis  fan 


Connecting  Rod 


°-  Guides 

•  Piston  Rod 

FIG.  329. 


The  total  pressure  is  the  pressure  per  square  inch  multiplied  by 
the  area  of  the  piston.  If  the  diameter  of  the  cylinder  is  20  inches, 
the  length  of  the  stroke  is  48  inches,  and  the  average  pressure  on 
one  side  of  the  piston  during  the  stroke  is  50  pounds  per  square 
inch,  the  total  work  of  the  steam  on  that  side  is 

314.16  X  50  X  4  =  62,832  foot-pounds. 

If  the  pressure  of  50  pounds  per  square  inch  is  measured  from 
the  atmospheric  pressure  as  the  zero  (as  is  customary)  and  if  the 
opposite  end  of  the  cylinder  is  entirely  open  during  the  entire 
stroke,  then  62,832  foot-pounds  represents  the  actual  work  done 
on  the  piston.  Usually,  there  is  some  back  pressure  in  the  oppo- 
site end  of  the  cylinder;  the  effective  work  on  the  piston  is, 
therefore,  somewhat  less  than  the  work  of  the  steam  on  one 

26 


402 


MECHANICS 


[ART.   241 


side.  The  actual  work  is  the  difference  between  the  positive 
work  in  the  steam  end  of  the  cylinder  and  the  negative  work  in 
the  other  end. 

The  average  pressure  in  an  engine  cylinder  is  determined  by 
means  of  an  indicator,  Fig.  330,  I.  An  indicator  is  a  pressure 
gage  which  is  connected  to  the  cylinder  by  means  of  a  short 
steam  pipe.  An  indicator  consists  essentially  of  a  small  cylinder 
in  which  moves  a  light  piston.  The  steam  pressure  in  the  cylin- 
der of  the  gage  is  balanced  by  a  calibrated  spring  on  the 
opposite  side  of  the  piston.  The  piston  actuates  a  light  lever, 
which  carries  a  pencil  at  the  end.  When  a  sheet  of  paper  is 


-Pipe  to 

Engine  Cylinder 


FIG.  330. 


moved  perpendicular  to  the  motion  of  the  pencil,  a  curve  whose 
ordinates  give  the  pressure  at  every  instant  is  drawn.  The  paper 
is  placed  on  a  drum  which  is  attached  to  the  engine  in  such  way 
that  its  motion  is  proportional  to  the  motion  of  the  piston. 

Figure  330,  II,  shows  an  indicator  card.  The  arrows  give  the 
direction  of  the  motion  of  the  pencil  relative  to  the  paper.  The 
distance  of  any  point,  such  as  G,  above  the  atmospheric  line 
measures  the  pressure  in  the  cylinder  of  the  engine  at  the  cor- 
responding point  of  its  stroke.  The  effective  force  on  the  piston 
is  the  difference  between  this  pressure  and  the  back  pressure  on 
the  opposite  side  of  the  piston.  This  back  pressure  may  be 
obtained  by  a  second  indicator  attached  to  the  opposite  end  of 
the  cylinder.  It  is  not  necessary  to  consider  together  the  pres- 
sure on  one  side  of  the  piston  and  the  back  pressure  on  the  other 
side  in  order  to  calculate  the  power.  The  effective  work  during 
one  cycle  may  be  calculated  from  the  difference  of  pressure  on  the 


CHAP.   XXVI] 


ENERGY  TRANSFER 


403 


same  side  of  the  cylinder.  In  Fig.  330,  II,  the  height  FG  of  the 
indicator  card  gives  the  difference  in  pressure  during  the  forward 
and  backward  strokes.  The  average  net  height  of  the  indicator 
card,  as  obtained  by  dividing  its  area  by  its  length,  gives  the 
mean  effective  pressure  for  the  one  end  of  the  cylinder. 

Problems 

1.  The  indicator  card  from  the  head  end  of  an  engine  cylinder  has  an  area 
of  8.40  square  inches,  and  the  card  from  the  crank  end  has  an  area  of  8.64 
square  inches.     Each  card  is  8  inches  long.     The  indicator  spring  used  in 
this  test  is  compressed  1  inch  by  a  pressure  of  40  pounds  per  square  inch  in 
the  cylinder  of  the  indicator.     Find  the  mean  effective  pressure. 

Ans.     m.e.p.  =  42  pounds  per  square  inch  in  the  head  end,  and 
43.2  pounds  per  square  inch  in   the  crank  end. 

2.  The  cylinder  of  the  engine  in  Problem  1  is  20  inches  in  diameter  and  the 
piston  rod  is  4  inches  in  diameter.     The  stroke  is  48  inches  and  the  speed  is 
80  revolutions  per  minute.     Find  the  indicated  horse  power. 

Ans.    284.5  hp. 


*••  Friction  Block 


FIG.  331. 


242.  Friction  Brake.— Fig.  331  shows  one  type  of  friction 
brake,  such  as  is  used  to  absorb  and  measure  the  power  developed 
by  an  engine  or  motor.  The  form  shown  consists  of  a  number 
of  wooden  friction  blocks  placed  around  the  rim  of  the  pulley. 
The  blocks  are  fastened  to  a  flexible  metal  band.  The  ends  of  the 
band  are  connected  together  by  bolts.  By  turning  the  nuts  on 
these  bolts  the  tension  may  be  varied  and  the  pressure  of  the 
blocks  on  the  pulley  may  be  adjusted  to  give  any  desired  value  to 
the  friction.  The  arm  B  ends  with  a  knife-edge  C  at  a  distance 
R  from  the  axis  of  the  pulley.  The  knife-edge  rests  on  a  vertical 


404 


MECHANICS 


[ART.   242 


support,  which  stands  on  a  platform  scale.  When  the  pulley  is 
turning  in  the  direction  of  the  arrow,  the  friction  of  the  blocks 
transmits  a  torque  to  the  brake.  This  torque  is  balanced  by  the 
upward  force  P  from  the  support  to  the  knife-edge. 

Suppose  that  the  pulley  is  held  stationary  and  that  a  force  P 
is  applied  to  the  knife-edge  at  right  angles  to  the  radius  R.  If 
the  torque  of  this  force  is  sufficient  to  balance  the  torque  of 
friction,  the  knife-edge  will  describe  a  circle  of  radius  R  about  the 
axis  of  the  pulley.  The  circumference  of  this  circle  is  2irR,  and 

the  work  done  by  the  force  P  during  one 

revolution  is 


-26 


U  =  ZirRP  =  2irT,  (I) 

in  which  T  is  the  torque. 

If  the  brake  is  held  stationary  by  the 
reaction  of  the  support  and  the  pulley 
is  rotated,  the  work  per  revolution  is  the 
same  as  it  would  be  if  the  pulley  were 
stationary  and  the  brake  were  rotated. 
In  all  cases,  the  work  per  revolution  = 
2irT. 

Problems 

1.  The  flywheel    of    an    engine    makes  300 
revolutions  per  minute.    A  friction  brake  on  the 
wheel  has  a  moment  arm  of  90  inches.     Find 
the  horsepower  when  the  reaction  at  the  end 
of  the  brake  arm  is  420  pounds. 

Ans.     178  hp. 

2.  The  arm  of  a  brake  which  is  attached  to 
the  pulley  of  a  motor  is  30  inches  long.     The 
motor  makes  1200  revolutions  per  minute.  Find 
the  horsepower  when  the  reaction  at  the  sup- 
port  is   60    pounds.      If  the  current  used  to 
drive  the  motor  is  130  amperes  at  220  volts, 
what  is  the  efficiency? 

Ans.     34.27  hp.;  89.3  per-cent. 

3.  The  pulley  of  Fig.  332  is   10  inches    in 
diameter  and  the  rope  is  %  inch  in  diameter.     The  total  weight  is  112 
pounds  and  the  spring  balance  reads  26  pounds.     Find  the  horsepower 
when  the  pulley  is  making  800  revolutions  per  minute.          Ans.     5.86  hp. 

4.  What  must  be  the  coefficient  of  friction  between  the  rope  and  the  pulley 
of  Problem  3? 

6.  The  flywheel  of  Problem  1  is  cooled  by  water  which  is  held  against  the 
rim  of  the  wheel  by  centrifugal  force.  The  water  is  supplied  to  the  wheel 
at  60  degrees  Fahrenheit  and  is  evaporated  at  210  degrees  Fahrenheit. 
How  many  pounds  of  water  are  required  per  hour? 


•/2/b. 


FIG.  332. 


CHAP.  XXVI] 


ENERGY  TRANSFER 


405 


243.  Cradle  Dynamometer. — The  output  of  a  small  motor 
may  be  measured  by  means  of  a  cradle  dynamometer.  The 
motor  is  mounted  on  a  frame  which  is  supported  by  a  pair  of 
knife-edges.  These  knife-edges  are  in  line  with  the  axis  of  the 
shaft  of  the  motor.  The  torque  of  the  belt  on  the  motor  tends  to 
rotate  the  frame  about  the  knife-edges.  Since  the  knife-edges  are 
in  line  with  the  axis  of  the  pulley,  the  direct  pull  of  the  belt  has  no 
effect  on  the  equilibrium.  The  torque  of  the  belt  is  balanced  by 
the  weights  P  and  the  poise  Q  on  a  horizontal  arm  attached  to  the 
frame.  When  the  motor  is  revolving  clockwise  with  the  belts 
extending  toward  the  left,  as  shown  in  Fig.  333,  the  tension  is 


Suspension 

Bar- 


'•-  Cradle 


FIG.  333 


greater  in  the  upper  belt  and  the  torque  from  the  belt  to  the 
armature  is  counter-clockwise.  This  torque  is  transmitted 
electromagnetically  from  the  armature  to  the  field,  and  then 
transmitted  mechanically  from  the  field  to  the  dynamometer. 

When  a  dynamo  which  is  driven  by  a  belt  is  to  be  tested, 
either  the  rotation  must  be  opposite  that  shown  in  Fig.  333,  or 
the  arm  which  carries  the  balancing  weights  must  extend  toward 
the  left. 

The  calculation  of  power  of  a  cradle  dynamometer  is  the  same 
as  the  calculation  for  a  friction  brake.  The  work  per  revolution 


244.  Transmission  Dynamometer.  —  A  transmission  dynamom- 
eter transmits  and  measures  power  without  dissipating  it  by 
transfer  into  heat.  Fig.  334  shows  one  type  of  a  dynamometer 
driven  by  a  belt.  The  pulley  B  is  driven  from  pulley  A  through 
the  dynamometer  pulleys  C  and  D.  The  dynamometer  is 
arranged  to  measure  the  pull  Q  in  the  part  of  the  belt  running 


406 


MECHANICS 


[ART.   244 


from  C  to  B  and  the  pull  P  in  the  part  of  the  belt  running  from  B 
to  D.  The  difference  P  —  Q  multiplied  by  the  radius  of  the 
pulley  B  (measured  to  the  center  of  the  belt)  gives  the  torque. 
Each  dynamometer  pulley  is  mounted  on  a  frame  which  is  sup- 
ported by  knife-edges.  The  knife-edges  which  support  pulley  C  are 
in  the  plane  of  the  center  of  the  belt  running  from  AtoC  and  the 
knife-edges  which  support  pulley  D  are  in  the  plane  of  the  center  of 


FIG.  334. 

the  belt  running  from  D  to  A .  The  tension  in  these  parts  of  the 
belt,  therefore,  exert  no  torque  on  the  frames  which  support  the 
dynamometer  pulleys.  The  moment  on  the  left  frame  is  the 
product  of  the  pull  Q  multiplied  by  the  diameter  of  pulley  C 
(from  center  to  center  of  belt):  and  the  moment  on  the  right 
frame  is  the  product  of  P  multiplied  by  the  diameter  of  pulley  D. 

Figure  335  shows  both  frames  attached  to  a  single  lever,  at 
equal  distances  on  opposite  sides  of  the  fulcrum  G.  The  moment 
on  this  lever  measures  the  difference  Q  —  P. 

It  is  not  absolutely  necessary  that  the  belt  should  run  vertical 
from  A  to  C  and  from  D  to  A.  The  belt  may  run  at  any  angle, 
provided  the  plane  of  the  center  passes  through  the  line  of  the 
knife-edges  on  each  side. 

Figure  336  shows  the  essentials  of  the  Robinson  dynamometer. 


CHAP.  XXVI] 


ENERGY  TRANSFER 


407 


The  spur  gear  A  drives  the  spur  gear  C  through  the  gear  B.     The 
gearwheel  B  is  mounted  on  a  frame  which  is  pivoted  at  0  in  the 


FIG.  335. 


rz^Bf-ri 


l::::iSp::::l 

X7"  LJl:  ""X 


FIG.  336. 


plane  tangent  to  the  pitch  circle  of  B.     The  reaction  between 
B  and  C  does  not,  therefore,  exert  any  moment  on  the  frame. 


408  MECHANICS  [ART.  245 

The  reaction  of  A  on  B  has  a  moment  arm  equal  to  the  diameter 
of  the  pitch  circle  of  B.  This  moment  is  balanced  by  the  weights 
at  the  end  of  the  lever  arm.  The  gearwheels  A  and  C  are 
mounted  on  a  frame.  Each  is  generally  connected  to  a  pulley 
by  which  the  power  is  transferred  to  a  belt. 

The  torque  in  a  shaft  may  be  measured  by  means  of  the  angle 
of  twist  of  a  portion  of  its  length.  The  angle  of  twist  in  a  shaft  of 
large  diameter  is  very  small.  Delicate  devices  have  been  made, 
however,  by  means  of  which  these  small  angles  may  be  measured 
in  a  rotating  shaft. 

Problem 

Devise  a  transmission  dynamometer  made  with  three  bevel-gearwheels. 

245.  Power  Transmission  by  Impact. — When  a  perfectly 
elastic  body  collides  with  another  perfectly  elastic  body,  there  is 
no  loss  of  energy.  If  the  final  velocity  of  one  of  the  bodies  is 
zero,  all  of  its  kinetic  energy  is  transferred  to  the  other  body. 


A  vrj 


FIG.  337. 

In  Figure  337,  A  is  a  relatively  large  body  which  is  moving 
with  a  velocity  vz,  when  it  is  struck  by  a  relatively  small  body 
of  mass  m  moving  in  the  same  direction  with  a  velocity  v\.  The 
change  of  velocity  of  m  during  compression  is  v\  —  v-*.  If  the 
bodies  are  perfectly  elastic,  the  change  during  compression  and 
restitution  is  2(vi  —  v^).  If  t  is  the  time  of  contact,  the  average 
force  is 


The  distance  which  the  mass  A  moves  during  the  time  t  is 
the  work  done  by  m  on  A  is,  therefore, 

2m  (PI  - 


Q 
The  kinetic  energy  of  the  mass  m  before  collision  was 


CHAP.  XXVI] 


ENERGY  TRANSFER 


409 


If  all  this  energy  is  transferred  to  A, 


0; 


(4) 
(5) 
(6) 


When  the  mass  of  A  is  so  large  relatively  that  its  velocity  is 
not  appreciably  changed  by  collision  with  m,  the  velocity  of  A 
must  be  one-half  as  great  as  the  initial  velocity  of  m  in  order  that 
all  the  energy  of  m  may  be  transferred.  The  change  of  velocity 


I 


Fio.  338. 


Fio.   339. 


of  m  during  compression  is  ^  and  the  change  during  restitution 

is  the  same;  the  final  velocity  of  m  is  zero  and  its  kinetic  energy 
is  zero.  V 

When  a  water  jet  strikes  a  plane  surface,  as  in  Fig.  338,  it 
behaves  almost  as  an  inelastic  body.  If  it  strikes  a  curved  sur- 
face tangentially,  it  is  deflected  with  little 
loss  of  energy.  If  the  surface  is  such  as 
to  deflect  the  jet  through  180  degrees,  as  in 
Fig.  339,  it  behaves  as  a  stream  of  perfectly 
elastic  particles. 

Figure  340  shows  a  section  of  one  bucket 
of  a  Pelton  impulse  wheel,  which  makes  use 
of  this  principle.  The  water  jet  strikes 
the  middle  of  the  bucket  and  divides.  Each  portion  of  the 
stream  has  its  direction  changed  nearly  180  degrees.  A  series  of 
such  buckets  is  arranged  around  the  rim  of  the  wheel  in  such  a 
way  that  the  jet  is  always  striking  one  or  two  buckets.  The 


FIG.  340. 


410  MECHANICS  [ART.  246 

maximum  efficiency  is  secured  when  the  velocity  of  the  buckets  is 
one-half  the  velocity  of  the  water  in  the  jet. 

Problems 

•  1.  A  Pelton  wheel  is  5  feet  in  diameter  from  center  to  center  of  buckets. 

It  is  driven  by  a  jet  which  has  a  velocity  of  120  feet  per  second.     What 

should  be  the  speed  of  the  wheel  for  maximum  efficiency?     Ans.     229  r.p.m. 

2.  What  should  be  the  linear  velocity  of  the  buckets  of  an  impulse  wheel 

which  is  driven  by  water  under  a  pressure  of  150  pounds  per  square  inch? 

246.  Summary. — Energy  is  measured  in  foot-pounds,  kilogram- 
meters,  gram  centimeters,  ergs,  and  joules. 

Power  is  rate  of  working.  A  horsepower  is  550  foot-pounds 
per  second.  A  watt  is  the  electrical  measure  of  power. 

746  watts  =  1  horsepower. 

The  relation  between  heat  energy  and  mechanical  energy  is 
called  the  mechanical  equivalent  of  heat. 

778  foot-pounds  =  1  British  thermal  unit. 
427  kilogram-meters  =  1  large  calorie. 

The  velocity  of  a  liquid  flowing  from  an  orifice  is  that  of  a  body 
which  falls  freely  from  a  height  equal  to  the  depth  of  liquid  above 
the  orifice. 

v  =  V2gh. 

The  quantity  is  the  velocity  multiplied  by  the  area  of  the  jet. 
For  an  orifice  in  a  thin  plate, 

Q  =  0.62A\/2^. 

The  work  of  a  water  jet  per  unit  of  time  is  the  product  of  the 
mass  per  unit  of  time  multiplied  by  the  height  of  the  water 
above  the  orifice. 

The  work  of  steam  in  a  cylinder  is  the  product  of  the  pressure 
multiplied  by  the  length  of  stroke.  The  pressure  is  the  area  of 
the  piston  multiplied  by  the  mean  effective  pressure.  The 
mean  effective  pressure  is  determined  from  the  average  height 
of  an  indicator  card. 

The  power  of  a  motor  may  be  measured  by  means  of  a  friction 
brake.  The  work  per  revolution  is 

U  =  27I-T, 


CHAP.  XXVI]  ENERGY  TRANSFER  411 

in  which  T  is  the  torque  about  the  axis  of  rotation.  The  torque 
of  small  motors  may  be  measured  by  means  of  a  cradle 
dynamometer. 

Transmission  dynamometers  measure  the  power  of  a  machine 
without  absorbing  it. 

When  perfectly  elastic  bodies  collide,  there  is  no  loss  of  energy. 
In  order  that  a  body  may  give  up  all  of  its  energy,  its 
final  velocity  must  be  zero.  When  a  continuous  stream  of 
particles  strikes  a  moving  body,  the  condition  of  maximum  effi- 
ciency requires  that  the  body  shall  be  moving  with  one-half  the 
velocity  of  the  particles.  A  jet  of  water  which  strikes  a  curved 
surface  tangentially  and  is  deflected  180  degrees  behaves  like  a 
body  which  is  highly  elastic. 


INDEX 


Absohite  units,  294,  322 
Acceleration,  linear,  264,  265,  266, 
267,  280,  342,  343 

angular,  354,  Chapter  XXIII 
Action  and  reaction,  32,  380 
Addition,  of  couples,  114 

of  numbers,  10 

of  vectors,  11,  12,  14,  15,  16,  17, 

18,  22,  23,  25 

Amplitude  of  vibration,  306 
Angle  of  friction,  192 
Angle  section,  moment  of  inertia  of, 

244,  245,  249 
Angular  motion,  354,  Chapter  XXI, 

368,  Chapter  XXIV. 
Atwood  machine,  282,  365 
Axis,  instantaneous,  339 

of  oscillation,  375 

of  suspension,  373 

transfer  of,  233,  248,  250,   258 


B 


Balance,  running,  347 

static,  346 
Ball  governor,  348 
Beam  balance,  221,  222,  223 
Bearing  friction,  195 
Belt  friction,  200 
Body,  rigid,  30 

free,  35,  36 

Bow's  method,  65,  99,  121 
Brake,  463 

British  thermal  unit,  397 
Bureau  of  Standards,  3 


Calculation  of  resultants  and  com- 

ponents,  44 
Calorie,  397 


Catenary,  129,  Chapter  VIII 
Center  of  gravity,  35,  151,  154  156, 

166 

by  integration,  159 
Center  of  mass,  35,  152 
Center  of  percussion,  391 
Center  of  pressure,   170,  252,  253, 

254 

Centimeter-gram-second  system,  295 
Central   forces,   327,    Chapter   XX 
Centrifugal  force,  344 
Channel  sections,  moment  of  inertia 

of,  245 
Chord,  96 
Circle,  moment  of  inertia  of,  242, 

243 

Circuit  of  force,  35 
Coefficient  of  discharge,  400 

of  friction,  190 
Collision  of  elastic  bodies,  383 

of  inelastic  bodies,  382 
Components  of  acceleration,  267 
of  force,  44 
of  a  vector,  12 
of  velocity,  264 
Compression,  29 
t   Concurrent,  coplanar  forces,  36,  43, 

Chapter  IV 
calculation    of    resultants    and 

components,  44 
equilibrium,  46,  48 
resultant,  37,  43 
trigonometric  solution,  53 
unknowns,  54 
(.    Concurrent  non-coplanar  forces,  141, 

Chapter  IX 

Cone,  moment  of  inertia  of,  231 
Cone  of  friction,  194 
Connecting  rod,  218,  401 
X Couples,  111,  Chapter  VI 
addition  of,  114,  116 
combination  with  forces,  116 
equivalent,  111 
413 


414 


INDEX 


Couples,  in  parallel  planes,  173 

as  vectors,  175 
Cradle  dynamometer,  405 
Cross-head,  218,  401 
Cubic  equation,  219 
Cylinder,    moment    of    inertia    of, 
229,  231 


Deflection  of  catenary,  130,  132 
Derrick,  68,  184 

Differential  appliances,  213,  214 
Dimensional  equations,  270,  294 
Direction  condition  of  equilibrium, 

92 

Direction  cosines,  21,  22 
Displacement,  angular,  334,  355 

linear,  259 

in  terms  of  velocity,  282 
Dynamometer,  405,  406 
Dyne,  7,  295 


E 


Effective  moment  arm,  56 
Efficiency,  207 
Elongation,  306 
Energy,  kinetic,  283 
potential,  207,  285 
transfer  of,  395,  Chapter  XXVI 
Equations,  concurrent,  coplanar 

forces,  49,  54,  55,  62,  68,  69 
concurrent,  non-coplanar  forces, 

148 

dimensional,  270 
non-concurrent,  coplanar  forces, 

87,  91,  92,  107 
non-concurrent,      non-coplanar 

forces,  180 

Equilibrant,  34,  46,  72 
Equilibrium,  31,  32,  40,  46,  53,  78 
calculation  by  moments,  60,  146 
calculation  by  resolutions,  48 
classes,  40,  220 
concurrent,  coplanar  forces,  46, 

47,  48 

concurrent,  non-coplanar  forces, 
144,  146,  147 


Equilibrium,    conditions  of   stable, 
79,  221 

of  couples,  115 

direction  condition  of,  92 

moment  condition  of,  60,  146 

non-concurrent,  coplanar  forces, 
86,  87,  115 

non-concurrent,      non-coplanar 
forces,  180 

of  parallel  forces,  78,  150 

surface  of,  224 

trigonometric   solution   for,   53 

by  work,  207 

Equipotential  surface,  224,  331 
Equivalent  mass,  357 
Erg,  295,  395 


Flexible  cord,  31,  129 
Fly-wheel  stresses,  351 
Force,  2,  3,  6,  29,  43,  273 

application  of,  29,  36 

centrifugal,  206 

moment  of,  56 

polygon  of,  46 

triangle  of,  38 
Force  circuit,  35 
Force   and   couple,    116,    118,    178 

and  motion,  273,  et  seq. 
Forces,    concurrent,    coplanar,    36, 
Chapter  IV 

concurrent,    non-coplanar,  141, 
Chapter  IX 

non-concurrent,     coplanar,  72, 
Chapter  V,  111,  120,  121 

non-concurrent,     non-coplanar, 
180 

resultant  of,  37,  42,   120,    122 
Formulas,  meaning  of,  7 
(Foot,  standard,  3,  7 
'Free  body,  35,  36 
Friction,  32,  33,  189,  Chapter  XII 
Friction  brake,  403 
Funicular  polygon,  121,  126,  127 

G 

Gaseous  state,  30 
Gee-pound,  297 


INDEX 


415 


Governor,  348,  349,  350 
Graphics  of  non-concurrent  forces, 
Chapter  VII 

of  vectors  in  space,  24 
Gravity,  acceleration  of,  275 

center  of,  35 

determination  of,  377 

motion  from,  286 
Gyration,  radius  of,  232 

H 

Harmonic  motion,  311,  318 
Heat  equivalent,  396 
Hinge,  34 
Horsepower,  396 


Impulse,  379 

Inclined  plane,  208,  216 

Independent  equations,  54,  55,   62, 

68,  69,  89,  91,  92,  107,  148, 

180 

Indicator,  402 
Inertia,  moment  of,  Chapters  XIV, 

XV 

product  of,  247 
Infinite  series,  135 
Input,  207 

Instantaneous  axis,  339 
Isochronous  vibration,  312 


Jointed  frame,  95,  103 
Joule,  395 

K 

Kilogram,  5,  6 
Kilogram-meter,  395 
Kinetic  energy,  283,  294,  307 
Kinetic  energy  of  rotation,  335,  336 
Kinetics,  1 


Length,  3,  4 
Lever,  215 


Link,  34 

Liquid  pressure,  169,  252 

Liquid  state,  30 

M 

Machines,  205,  Chapter  XIII 
Maclaurin's  Theorem,  135,  136 
Mass,  2,  3,  4,  5 

relation  to  weight,  7 
engineer's  unit  of,  297 
Matter,  4 

Maximum  moment  of  inertia,  251 
Mechanical  advantage,  207,  226 
Mechanical  equivalent  of  heat,  396 
Mechanics,  1 
Meter,  standard,  3 
Moment,  56 

of  couple,  111 
of  non-coplanar  forces,  145 
X  resultant,  58 
Moment  of  inertia,   228,   Chapters 

XIV,  XV 

of  connected  bodies,  238 
maximum,  251 

of  plane  area,  241,  Chapter  XV 
of  plate,  235 
polar,  241 
transfer  of,  233 
Moment  of  momentum,  386 
Momentum,  379,  Chapter  XXV 
Motion,  259,  Chapter  XVI 
Motion   and   force,    273,    Chapters 
XVII,  XVIII 


N 


X  Neutral  equilibrium,  42 

Newton's  Laws  of  Motion,  6 
)(    Non-concurrent  forces,  37,  72,  111 
Non-coplanar  forces,  37 
Numbers,  9 


Origin  of  moments,  57 

of  a  vector,  11 
Oscillation,  axis  of,  375 
Output,  207 


416 


INDEX 


Panel,  96 

Parabola,  center  of  gravity  of,  165  ' 
Parabola  of  flexible  cord,  137 
Parallel  forces,  72,  74,  75,  78,  124, 

149 
Parallelepiped,    moment   of   inertia 

of,  230 

Pelton  wheel,  409 
Pendulum,  gravity,  373 

simple,  374 

torsion,  369 
Percussion,  axis  of,  391 
Plane,  smooth,  33       .. 
Plate  and  angle  sections,  245 
Plate  elements,  236 
Polar  moments  of  inertia,  241 
Pole,  121 
Polygon,  of  forces,  46,  55 

funicular,  121,  126,  127 
Potential  energy,  207,  304 
Pound  force,  7,  8,  294,  296 
Pound  mass,  5,  6,  8 
Poundal,  7,  296,  299 
Power,  396,  399 

Power  transmission  by  impact,  408 
Pressure,  center  of,   170,  252,  253, 
254 

of  liquid,  169 

Product   of   inertia,    247,    248,    249 
Product  of  vectors,  25,  26 
Projectiles,  287 
Pulleys,  211 

Q 

Quantity,  9 

represented  by  lines,  9,  10 
represented  by  numbers,  9 
fundamental,  2 


Radius  of  gyration,  232 

Range,  290 

Ray,  121 

Reactions,  parallel,  124 

non-parallel,  127 

of  supports,  358,  360,  362 


Reference  circle,  311 
Resolution,  of  vectors,  13,  16 

of  forces,  38,  141 
Resultant,  34 

of  concurrent  forces,  37,  44,  141, 
142 

of  couples,  176 

of  non-concurrent  forces,  81,  82 

of  non-parallel  forces  graphic- 
ally, 81,  122 

of  parallel  forces,  72,  76 
Robinson  dynanometer,  407 
Roller  bearings,  200 
Rolling  friction,  198 
Rotation,  335 

Rotation  and  translation,  337,  338 
Rowland's  experiments,  397 


Scalar  product,  26,  27 
Screw,  213 

differential,  214 
Second,  4 

Sections,  method  of,  100 
Sensibility  of  balance,  222 
Shear,  29 

Simple  harmonic  motion,  311 
Slugg,  297 
Smooth  hinge,  34 
Smooth  surface,  32,  33 
Solid,  29 
Space,  2,  3 

Sphere,  moment  of  inertia  of,  230 
Spring,  energy  of ,  304,  307 

time  of  vibration,  309 
Stable  equilibrium,  41,  79,  221 
Standard  pound  force,  8,  294,  296 
Standards  of  mass,  etc.,  35 
Starting  friction,  33,  189 
Statics,  1,  2 

Strength  of  materials,  71 
String  polygon,  121 
Subtraction,  of  numbers,  10 

of  vectors,  18,  19 
Suspension,  axis  of  373 


Tension,  29 
Terminus  of  vector,  11 


INDEX 


417 


Three  force  member,  103 

Time,  234 

Time  of  vibration,  309,  370 

Toggle  joint,  217 

Transfer  of  axes,  233,  248,  250,  251 


Vectors,  components,  12,  13,  22,  24, 

25 

definition,  10 
resolution,  13 
subtraction,  18,  19 


Translation  and  rotation,  337,  338      Vectors  in  space,  19,  20,  21,  22,  23, 


Transmission  dynamometer,  405 
Triangle  of  forces,  45 
Trigonometric  solution,  53 
Trusses,  95,  96,  97,  98 

by  method  of  sections,  100 
Two  force  member,  103 


U 


24,  25 

graphical  resolution  of,  24,  25 
Velocity,  angular,  334 
Velocity,  linear,  260,  261,  262,  305 

average,    262 

as  a  derivative,  263 
Vibration,  307,  368 
Virtual  work,  215,  217 


Units,    systems    of,    299,    Chapter 

XVIII 
Unknowns,  54 

concurrent,  coplanar  forces,  54      Water  Jet> 


W 


concurrent,  non-coplanar  forces, 


Watt> 


148  Weight,  6 
non-concurrent,  coplanar  forces,  relation  to  mass,  7 

g7  variation  with  latitude,  8 

non-concurrent,      non-coplanar  Wheel  and  axle>  21° 
forces,  180  differential,  214 

X  Unstable  equilibrium,   40,   41,   221  Wmdlass»  185 

Work,  39,  205,  328 

v  virtual,  215,  217 


Vector  products,  25,  26 
Vectors,  addition  of,  11,  14,  15,  16, 
17,  18,  22,  23,  25 


Yard,  standard,  3 


'27 


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